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Quadratic Equations
Standard Form and Identification
A quadratic equation is any equation that can be written in standard form, where the highest power of the variable is 2. The word "quadratic" comes from the Latin word for "square" — the variable is literally being squared.
$$ax^2 + bx + c = 0, \quad a \neq 0$$
Every quadratic equation can be rearranged into this three-term form, where all terms sit on one side and equal zero.
$a$=Coefficient of $x^2$ — must be non-zero for the equation to be quadratic(unitless)
$b$=Coefficient of $x$ — the linear term(unitless)
$c$=Constant term — the number with no $x$ attached(unitless)
$x$=The variable (unknown) we are solving for(unitless)
$a = 0$
→Not quadratic — degrades to a linear equation $bx + c = 0$.
$b = 0$
→Pure quadratic: $ax^2 + c = 0$, solvable by isolating $x^2$.
$c = 0$
→Factored form $x(ax + b) = 0$ — one root is always $x = 0$.
Three-Term Structure: Look for exactly three terms when rearranged: an $x^2$ term, an $x$ term, and a constant. Missing terms simply have a coefficient of zero.
Rearrange First: Equations like $3x^2 = 7x - 2$ are quadratic but not in standard form. Move everything to one side: $3x^2 - 7x + 2 = 0$.
Discriminant Dependence: The values of $a$, $b$, and $c$ alone determine how many real answers (roots) the equation has — this is the power of standard form.
The roots of a quadratic equation are the values of $x$ that satisfy the equation — plugging them in makes the left side equal zero. Geometrically, roots are the $x$-coordinates where the parabola crosses the $x$-axis.
$$\text{If } \alpha \text{ is a root, then } a\alpha^2 + b\alpha + c = 0$$
A root is an $x$-value that makes the quadratic expression evaluate to exactly zero.
$\alpha, \beta$=The two roots (solutions) of the quadratic equation(unitless)
Two distinct roots
→The parabola crosses the $x$-axis at two different points.
One repeated root
→The parabola touches the $x$-axis at exactly one point (the vertex sits on the axis).
No real roots
→The parabola never reaches the $x$-axis — it floats entirely above or below.
Maximum Two Roots: A quadratic equation has at most two real roots. It can never have three or more.
Verification Shortcut: To check whether a number is a root, substitute it directly. If the expression equals zero, it is a valid root.
Zero-Product Link: When you factor the quadratic, each factor set to zero gives a root — this is the core logic behind the factoring method.
Solving by Factorization
Factorization is the fastest way to solve a quadratic — if it works. The idea is to split the middle term so the expression becomes a product of two brackets, then use the zero product property to find the roots.
$$(px + q)(rx + s) = 0 \implies x = -\frac{q}{p} \text{ or } x = -\frac{s}{r}$$
When a product of two factors equals zero, at least one factor must be zero — each gives a root.
$(px + q)$=First linear factor(unitless)
$(rx + s)$=Second linear factor(unitless)
$x = -q/p$=Root from the first factor(unitless)
$x = -s/r$=Root from the second factor(unitless)
AC Method: Multiply $a \times c$, then find two numbers that multiply to $ac$ AND add to $b$. Split the middle term with these numbers, then factor by grouping.
Rearrange and Factor: Always start in standard form. For $x^2 - 5x + 6 = 0$: find $-2$ and $-3$ (multiply to 6, add to $-5$), then $(x-2)(x-3) = 0$.
When $a = 1$: This is the easiest case. Find two numbers with product $c$ and sum $b$.
When $a \neq 1$: Use the AC method: for $2x^2 + 7x + 3 = 0$, $ac = 6$, split into $2x^2 + 6x + x + 3$, then factor as $(2x + 1)(x + 3) = 0$.
A perfect square trinomial is a quadratic that factors into the square of a binomial. Recognizing these saves time — the two roots are identical (a repeated root).
$$x^2 + 2kx + k^2 = (x + k)^2 = 0 \implies x = -k \text{ (double root)}$$
When the quadratic is a perfect square, the middle term is exactly twice the product of the square root of the first and last terms.
$k$=Half of the $x$ coefficient — the number inside the squared bracket(unitless)
$x^2 + 2kx + k^2$
→Factors as $(x + k)^2$ with a double root at $x = -k$.
$x^2 - 2kx + k^2$
→Factors as $(x - k)^2$ with a double root at $x = k$.
Quick Check: If the first term is $x^2$ and the last term is a perfect square, check whether the middle term equals $2\sqrt{\text{first} \times \text{last}}$.
Discriminant Is Zero: A perfect square trinomial always has discriminant $\Delta = 0$, confirming exactly one repeated root.
Completing the Square: Any quadratic can be converted to a perfect square trinomial by adding and subtracting the right constant — this is the completing-the-square technique.
The Quadratic Formula
The quadratic formula is the universal solver — it works for every quadratic equation regardless of whether it factors cleanly. Derive it once (via completing the square), then use it as your reliable fallback for any quadratic.
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The two roots are obtained by adding and subtracting the square root term — the $\pm$ produces both values in a single formula.
$-b$=Negated linear coefficient(unitless)
$\sqrt{b^2 - 4ac}$=The discriminant's square root — determines how far the roots are from $-b/2a$(unitless)
$2a$=Denominator — twice the $x^2$ coefficient(unitless)
$\sqrt{b^2 - 4ac} = 0$
→Both roots are the same: $x = -b/2a$ (double root).
$\sqrt{b^2 - 4ac} > 0$
→Two distinct real roots.
$b^2 - 4ac < 0$
→No real roots — the square root of a negative number is not a real number.
Why It Always Works: The formula is derived by completing the square on $ax^2 + bx + c = 0$, which is always algebraically possible.
Numerator Structure: $-b$ is the center, and $\pm\sqrt{b^2-4ac}$ is the spread. The roots are symmetric around $-b/2a$.
Denominator Matters: Always divide the entire numerator by $2a$. A common slip is dividing only one term.
Irrational Roots: When the discriminant is not a perfect square, the roots involve $\sqrt{}$ and are irrational — leave them in simplest radical form unless a decimal is requested.
When to Use the Formula vs Factorization
•
Factorization is faster when $a = 1$ and $c$ has few factor pairs
•
Use the formula when factorization is messy or when roots are irrational
•
Use the formula when you also need the discriminant value (e.g., to classify roots)
•
For competitive questions, the formula is safer — less room for arithmetic error
Nature of Roots (Discriminant)
The discriminant is the expression under the square root in the quadratic formula. Without solving the equation, it tells you exactly how many real roots exist and what they look like.
$$\Delta = b^2 - 4ac$$
The discriminant determines the nature of the roots by the sign and size of $b^2 - 4ac$.
$\Delta$=Discriminant of the quadratic equation(unitless)
$a, b, c$=Coefficients from the standard form $ax^2 + bx + c = 0$(unitless)
$\Delta > 0$ and a perfect square
→Two distinct rational roots — the equation factors cleanly.
$\Delta > 0$ and not a perfect square
→Two distinct irrational roots — the formula produces a $\sqrt{}$ term.
$\Delta = 0$
→One repeated real root — the parabola touches the $x$-axis at exactly one point.
$\Delta < 0$
→No real roots — the parabola does not intersect the $x$-axis.
Sign Is Everything: The sign of $\Delta$ tells you the count of real roots; its value tells you how far apart they are.
Perfect Square Shortcut: If $\Delta$ is a perfect square (like 25 or 49), the roots are rational — factorization should work.
Irrelevant Root Values: The discriminant tells you the nature of roots but not their actual values. To find the values, use the full formula.
Condition Questions: Problems like "find $k$ so that the equation has equal roots" mean: set $\Delta = 0$ and solve for $k$.
Discriminant Quick-Reference Table
•
$\Delta > 0$, perfect square → 2 rational roots (factorable)
•
$\Delta > 0$, not perfect square → 2 irrational roots (use formula)
•
$\Delta = 0$ → 1 repeated root ($x = -b/2a$)
•
$\Delta < 0$ → 0 real roots
Sum and Product of Roots
Without solving the equation, Vieta's formulas give you the sum and product of the roots directly from the coefficients. These relationships are deeply useful for constructing equations, checking answers, and solving problems where only root properties are given.
$$\alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a}$$
The sum of roots equals negative $b$ over $a$; the product equals $c$ over $a$.
$\alpha + \beta$=Sum of the two roots(unitless)
$\alpha\beta$=Product of the two roots(unitless)
$a, b, c$=Coefficients from $ax^2 + bx + c = 0$(unitless)
$\alpha + \beta = 0$
→Roots are opposites (e.g., 3 and $-3$). This means $b = 0$.
$\alpha\beta = -1$
→Roots are reciprocals with opposite signs (e.g., 2 and $-1/2$). Here $c/a = -1$.
$\alpha\beta = 1$
→Roots are reciprocals (e.g., 2 and $1/2$). Here $c = a$.
Sign Flip for Sum: The negative sign in $-b/a$ is the most common error source. If roots are both positive, $b$ must be negative (and vice versa).
Product Tells Both Signs: If $c/a > 0$, roots have the same sign (both positive or both negative). If $c/a < 0$, roots have opposite signs.
Quick Answer Check: After finding roots, verify: their sum should equal $-b/a$ and their product should equal $c/a$.
Derivation Insight: These formulas come from expanding $(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta = 0$ and matching coefficients.
Given the roots (or their sum and product), you can reconstruct the original quadratic equation by reversing Vieta's formulas. This is the backbone of problems that say "form an equation whose roots are...".
$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$
The required equation is built by placing the negative sum as the $x$ coefficient and the product as the constant term.
$x^2$=The quadratic term — always 1 when $a = 1$(unitless)
$-(\alpha + \beta)$=The $x$ coefficient — negative of the root sum(unitless)
$\alpha\beta$=The constant term — the root product(unitless)
Given Roots Directly: If roots are $3$ and $-2$, then sum $= 1$, product $= -6$, so the equation is $x^2 - x - 6 = 0$.
Given Modified Roots: For roots that are $\alpha + 2$ and $\beta + 2$, compute the new sum $(\alpha + \beta + 4)$ and new product, then plug into the formula.
Given Sum and Product Only: Even without individual roots, the equation is determined — $\alpha + \beta$ and $\alpha\beta$ are enough.
Non-Monic Equations: If the leading coefficient must be $a$ (not $1$), multiply through: $ax^2 - a(\alpha + \beta)x + a\alpha\beta = 0$.
Word Problems and Applications
Many real-world situations — profit optimization, area calculations, speed-distance-time, and investment returns — produce quadratic equations. The skill is translating words into the standard form $ax^2 + bx + c = 0$ and then selecting the root that makes physical sense.
$$\text{Real-world scenario} \xrightarrow{\text{translate}} ax^2 + bx + c = 0 \xrightarrow{\text{solve}} \alpha, \beta \xrightarrow{\text{filter}} \text{valid answer}$$
Set up the equation from the word problem, solve it, and then reject any root that violates the real-world constraints (like negative length or negative time).
$\alpha, \beta$=Two mathematical roots of the equation(depends on context)
Define the Variable Clearly: Always state what $x$ represents before writing the equation. If $x$ is the width of a rectangle, say so explicitly.
Two Answers, One Valid: Most word problems yield two roots. One usually violates a constraint (negative length, time before departure, etc.) — reject it and state why.
Common Setups: Area = length $\times$ width, profit = revenue $-$ cost, $d = s \times t$ with two legs, and $P(1 + r)^n$ for compound scenarios.
Check the Question: Does it ask for the value of $x$, or for something derived from $x$ (like the area, or the other dimension)? Read carefully.
Area and dimension problems are the most common business-application quadratics. When you know the total area and a relationship between length and width, the substitution produces a quadratic in one variable.
$$\text{Area} = l \times w \implies \text{Substitute } l = f(w) \implies \text{Quadratic in } w$$
Express one dimension in terms of the other using the given relationship, substitute into the area formula, and solve the resulting quadratic.
$l$=Length of the rectangle(length units)
$w$=Width of the rectangle(length units)
Length-Width Relationship: Common phrasing: "length exceeds width by 5" ($l = w + 5$), "length is twice the width" ($l = 2w$), or "perimeter is 40" ($2(l + w) = 40$).
Rejecting Negative Dimensions: A rectangle cannot have a negative side. Always discard negative roots.
Square Problems: When length $=$ width, $s^2 = \text{area}$ is not quadratic but square-root based. It becomes quadratic when the shape involves a border or margin.
Plot Problems: The area of a rectangular plot with a path/border around it is a common setup — the inner and outer rectangles share a dimension relationship.
Application Strategy
Worked Examples
Problem Statement
A small business's monthly profit (in thousands of rupees) is modeled by $P = x^2 - 7x + 10$, where $x$ is the number of employees beyond a base staff. For how many additional employees does the business break even (profit = 0)?
factorizationzero product property
Strategic Toolkit
$$x^2 - 7x + 10 = 0$$
Step 1
Set profit to zero
To find break-even points, set the profit expression equal to zero.
$$x^2 - 7x + 10 = 0$$
Step 2
Find two numbers
We need two numbers that multiply to $10$ and add to $-7$. The numbers are $-2$ and $-5$.
$$(-2) \times (-5) = 10, \quad (-2) + (-5) = -7$$
Step 3
Factor the quadratic
Write the quadratic as a product of two binomials using the numbers found.
$$(x - 2)(x - 5) = 0$$
Step 4
Apply zero product property
Set each factor equal to zero to find both roots.
$$x - 2 = 0 \implies x = 2 \quad | \quad x - 5 = 0 \implies x = 5$$
Step 5
Interpret the results
The business breaks even at 2 or 5 additional employees beyond the base staff.
$$\text{Break-even: } x = 2 \text{ or } x = 5$$
Problem Statement
A trader's daily revenue follows the model $R = -2x^2 + 12x + 14$ where $x$ is the number of hours spent at the market. Find the hours at which the trader stops earning (revenue = 0).
quadratic formulastandard form
Strategic Toolkit
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Step 1
Identify the coefficients
Rewrite in standard form $ax^2 + bx + c = 0$ and extract $a$, $b$, $c$.
$$-2x^2 + 12x + 14 = 0 \implies a = -2,\; b = 12,\; c = 14$$
Step 2
Compute the discriminant
Calculate $b^2 - 4ac$ to check the nature of the roots.
$$\Delta = 12^2 - 4(-2)(14) = 144 + 112 = 256$$
Step 3
Apply the quadratic formula
Substitute $a$, $b$, and $\sqrt{\Delta}$ into the formula.
$$x = \frac{-12 \pm \sqrt{256}}{2(-2)} = \frac{-12 \pm 16}{-4}$$
Step 4
Calculate both roots
Evaluate each case: with $+$ and with $-$.
$$x = \frac{-12 + 16}{-4} = \frac{4}{-4} = -1 \quad | \quad x = \frac{-12 - 16}{-4} = \frac{-28}{-4} = 7$$
Step 5
Reject the invalid root
Negative hours make no physical sense, so discard $x = -1$.
$$\text{Valid answer: } x = 7 \text{ hours}$$
Problem Statement
For the equation $3x^2 - 5x + 2 = 0$, determine the nature of the roots without solving.
discriminant
Strategic Toolkit
$$\Delta = b^2 - 4ac$$
Step 1
Identify coefficients
Extract $a$, $b$, and $c$ from the standard form.
$$a = 3,\; b = -5,\; c = 2$$
Step 2
Compute the discriminant
Plug values into $b^2 - 4ac$.
$$\Delta = (-5)^2 - 4(3)(2) = 25 - 24 = 1$$
Step 3
Interpret the result
Since $\Delta = 1 > 0$ and $1$ is a perfect square, there are two distinct rational roots.
$$\Delta = 1 > 0 \implies \text{two distinct rational roots}$$
Problem Statement
Without solving $2x^2 + 8x - 15 = 0$, find the sum and product of its roots.
Vieta's formulas
Strategic Toolkit
$$\alpha + \beta = -\frac{b}{a} \quad | \quad \alpha\beta = \frac{c}{a}$$
Step 1
Identify coefficients
Extract $a$, $b$, and $c$ from the equation.
$$a = 2,\; b = 8,\; c = -15$$
Step 2
Find the sum of roots
Apply Vieta's sum formula: negative $b$ over $a$.
$$\alpha + \beta = -\frac{8}{2} = -4$$
Step 3
Find the product of roots
Apply Vieta's product formula: $c$ over $a$.
$$\alpha\beta = \frac{-15}{2} = -7.5$$
Problem Statement
Form a quadratic equation whose roots are $\frac{2}{3}$ and $-4$.
Vieta's formulasforming equations
Strategic Toolkit
$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$
Step 1
Compute the sum of roots
Add the two given roots.
$$\alpha + \beta = \frac{2}{3} + (-4) = \frac{2}{3} - \frac{12}{3} = -\frac{10}{3}$$
Step 2
Compute the product of roots
Multiply the two given roots.
$$\alpha\beta = \frac{2}{3} \times (-4) = -\frac{8}{3}$$
Step 3
Build the equation
Substitute the sum and product into Vieta's form, then clear fractions by multiplying by 3.
$$x^2 - \left(-\frac{10}{3}\right)x + \left(-\frac{8}{3}\right) = 0 \implies x^2 + \frac{10}{3}x - \frac{8}{3} = 0$$
Step 4
Clear fractions
Multiply every term by 3 to get integer coefficients.
$$3x^2 + 10x - 8 = 0$$
Problem Statement
A rectangular garden has an area of 135 m². Its length is 6 m more than its width. Find the dimensions of the garden.
word problemsfactorization
Strategic Toolkit
$$w(w + 6) = 135$$
Step 1
Define the variable
Let $w$ be the width in meters. Then the length is $w + 6$ meters.
$$\text{Width} = w, \quad \text{Length} = w + 6$$
Step 2
Set up the area equation
Area equals length times width, so substitute and set equal to 135.
$$w(w + 6) = 135 \implies w^2 + 6w - 135 = 0$$
Step 3
Factor the quadratic
Find two numbers that multiply to $-135$ and add to $6$. The numbers are $15$ and $-9$.
$$(w + 15)(w - 9) = 0$$
Step 4
Solve and filter
Setting each factor to zero gives $w = -15$ or $w = 9$. Reject the negative width.
$$w = 9 \text{ m}, \quad l = w + 6 = 15 \text{ m}$$
Step 5
Verify
Check that the dimensions give the correct area.
$$9 \times 15 = 135 \text{ m}^2 \checkmark$$
Problem Statement
Find the value of $k$ for which the equation $(k + 1)x^2 - 6(k + 1)x + 3(k + 9) = 0$ has equal roots.
discriminant
Strategic Toolkit
$$\Delta = b^2 - 4ac = 0$$
Step 1
Identify coefficients
Extract $a$, $b$, and $c$ in terms of $k$.
$$a = k + 1,\; b = -6(k + 1),\; c = 3(k + 9)$$
Step 2
Set up the discriminant condition
For equal roots, set $\Delta = 0$.
$$\Delta = [-6(k+1)]^2 - 4(k+1)[3(k+9)] = 0$$
Step 3
Simplify the equation
Expand and simplify the discriminant expression. Note that $k + 1$ is a common factor.
$$36(k+1)^2 - 12(k+1)(k+9) = 0$$
Step 4
Factor out common terms
Factor out $12(k + 1)$ from both terms.
$$12(k+1)[3(k+1) - (k+9)] = 0$$
Step 5
Solve for $k$
The factor $12(k+1) = 0$ gives $k = -1$, but this makes $a = 0$ (not quadratic). Solve the second factor for the valid answer.
$$3(k+1) - (k+9) = 0 \implies 3k + 3 - k - 9 = 0 \implies 2k = 6 \implies k = 3$$
Problem Statement
A ball is thrown upward and its height after $t$ seconds is $h = -5t^2 + 20t + 15$ meters. After how many seconds does the ball hit the ground?
quadratic formulaword problems
Strategic Toolkit
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Step 1
Set height to zero
The ball hits the ground when $h = 0$.
$$-5t^2 + 20t + 15 = 0 \implies 5t^2 - 20t - 15 = 0 \implies t^2 - 4t - 3 = 0$$
Step 2
Identify simplified coefficients
After dividing by 5, the simplified equation has $a = 1$, $b = -4$, $c = -3$.
$$a = 1,\; b = -4,\; c = -3$$
Step 3
Apply the quadratic formula
Substitute into the formula with simplified coefficients.
$$t = \frac{4 \pm \sqrt{16 + 12}}{2} = \frac{4 \pm \sqrt{28}}{2} = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$$
Step 4
Select the valid root
Since $\sqrt{7} \approx 2.65$, the two roots are $t \approx -0.65$ and $t \approx 4.65$. Reject the negative time.
$$t = 2 + \sqrt{7} \approx 4.65 \text{ seconds}$$
Memorize formulas
Quadratic Formula
Use when factorization is difficult or roots are irrational
Formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Discriminant
Use to determine the nature of roots without solving
Formula
$\Delta = b^2 - 4ac$
Sum of Roots
Use to find or verify the sum of roots directly from coefficients
Formula
$\alpha + \beta = -\frac{b}{a}$
Product of Roots
Use to find or verify the product of roots directly from coefficients
Formula
$\alpha\beta = \frac{c}{a}$
Equation from Roots
Use to construct a quadratic when given its roots or their sum and product
Formula
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Standard Form
Use as the canonical form for identifying coefficients and applying formulas
Formula
$ax^2 + bx + c = 0, \quad a \neq 0$
Review
1
Which of the following is a quadratic equation in standard form $ax^2 + bx + c = 0$ where $a \neq 0$?
Ans
$3x^2 - 5x + 2 = 0$
Review
Tap to reveal
Check each equation against the standard form criteria
The equation must be polynomial with highest degree 2, and the coefficient of $x^2$ must be non-zero.
$$3x^2 - 5x + 2 = 0 \implies a = 3,\ b = -5,\ c = 2 \quad \checkmark$$
2
A student's practice score is modeled by $-x^2 + 10x$ where $x$ is the number of hours studied. If the score is 21 points, how many hours could the student have studied?
Ans
3 or 7 hours
Review
Tap to reveal
Set up the quadratic equation
Equate the score model to 21 and rearrange to standard form.
$$-x^2 + 10x = 21 \implies x^2 - 10x + 21 = 0$$
Factor and solve
Find two numbers that multiply to 21 and add to $-10$.
$$(x - 3)(x - 7) = 0 \implies x = 3 \text{ or } x = 7$$
3
Using the quadratic formula, solve $2x^2 + 3x - 2 = 0$.
Ans
$x = \frac{1}{2}$ and $x = -2$
Review
Tap to reveal
Calculate the discriminant
Compute $\Delta = b^2 - 4ac$ for the given equation.
$$\Delta = 3^2 - 4(2)(-2) = 9 + 16 = 25$$
Apply the quadratic formula
Substitute $a$, $b$, and $\Delta$ into $x = \frac{-b \pm \sqrt{\Delta}}{2a}$.
$$x = \frac{-3 \pm 5}{4} \implies x = \frac{1}{2} \text{ or } x = -2$$
4
For the equation $x^2 - 6x + 9 = 0$, what is the nature of its roots?
Ans
Two equal real roots
Review
Tap to reveal
Compute the discriminant
Use $\Delta = b^2 - 4ac$ with $a = 1$, $b = -6$, $c = 9$.
$$\Delta = 36 - 36 = 0$$
Classify the roots
When $\Delta = 0$, the quadratic has exactly one real root repeated twice.
$$(x - 3)^2 = 0 \implies x = 3 \text{ (double root)}$$
5
If the roots of $2x^2 - 8x + 3 = 0$ are $\alpha$ and $\beta$, what is $\alpha + \beta$?
Ans
$4$
Review
Tap to reveal
Identify coefficients and apply Vieta's formula
For $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$.
$$\alpha + \beta = -\frac{-8}{2} = 4$$
6
A rectangular garden has a length that is 3 meters more than its width. If the area of the garden is 70 m², find the width.
Ans
7 m
Review
Tap to reveal
Set up the quadratic from the area condition
Express the area as a product of width and length, then rearrange to standard form.
$$w(w + 3) = 70 \implies w^2 + 3w - 70 = 0$$
Solve by factorization
Find two numbers that multiply to $-70$ and add to $3$.
$$(w + 10)(w - 7) = 0 \implies w = 7 \text{ m}$$
7
A painter can paint a wall in $x$ hours and their apprentice can paint it in $(x + 6)$ hours. Working together they finish in 4 hours. The combined rate equation simplifies to $x^2 - 2x - 24 = 0$. What is the value of $x$?
Ans
$x = 6$
Review
Tap to reveal
Factor the quadratic equation
Find two numbers that multiply to $-24$ and add to $-2$.
$$(x - 6)(x + 4) = 0$$
Select the valid solution
Reject the negative root since time must be positive.
$$x = 6 \text{ hours}$$
8
Solve $x^2 - 4x - 1 = 0$ using the quadratic formula.
Ans
$x = 2 \pm \sqrt{5}$
Review
Tap to reveal
Calculate the discriminant
$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20$.
$$\sqrt{\Delta} = \sqrt{20} = 2\sqrt{5}$$
Apply the quadratic formula and simplify
Substitute into $x = \frac{-b \pm \sqrt{\Delta}}{2a}$ and reduce.
$$x = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$$
9
An architect designs a rectangular window where the length is $x$ cm and the width is $(x - 4)$ cm. For an area of 32 cm², the equation is $x^2 - 4x - 32 = 0$. What is the discriminant, and what does it tell us about the roots?
Ans
$\Delta = 144$; two distinct real roots
Review
Tap to reveal
Compute the discriminant
Use $\Delta = b^2 - 4ac$ with $a = 1$, $b = -4$, $c = -32$.
$$\Delta = 16 - 4(1)(-32) = 16 + 128 = 144$$
Interpret the discriminant
$\Delta = 144 > 0$ means two distinct real roots exist.
$$x = \frac{4 \pm 12}{2} \implies x = 8 \text{ or } x = -4$$
10
The roots of $3x^2 + 5x - 7 = 0$ are $\alpha$ and $\beta$. What is $\alpha\beta$?
Ans
$-\frac{7}{3}$
Review
Tap to reveal
Apply Vieta's formula for product of roots
For $ax^2 + bx + c = 0$, the product of roots is $\frac{c}{a}$.
$$\alpha\beta = \frac{-7}{3}$$
11
A company's monthly revenue $R$ (in thousands of rupees) from selling $x$ units is $R = -x^2 + 40x$. How many units must be sold to maximize revenue?
Ans
20 units
Review
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Identify the vertex of the parabola
For $R = ax^2 + bx$, the maximum occurs at $x = -\frac{b}{2a}$ when $a < 0$.
$$x = -\frac{40}{2(-1)} = 20$$
Verify the maximum revenue
Substitute $x = 20$ back into the revenue function.
$$R = -(20)^2 + 40(20) = -400 + 800 = 400 \text{ thousand rupees}$$
12
A quadratic equation has roots $3$ and $-5$. Which of the following is the correct equation?
Ans
$x^2 + 2x - 15 = 0$
Review
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Calculate sum and product of the given roots
Use Vieta's relationships in reverse to find $\alpha + \beta$ and $\alpha\beta$.
$$\alpha + \beta = -2, \quad \alpha\beta = -15$$
Form the equation
Substitute into $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
$$x^2 + 2x - 15 = 0$$
13
Find $k$ such that $(k+1)x^2 - 6(k+1)x + 3(k+9) = 0$ has equal roots.
Ans
$k = 3$
Review
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Set the discriminant to zero
With $a = k+1$, $b = -6(k+1)$, $c = 3(k+9)$, compute $b^2 - 4ac = 0$.
$$36(k+1)^2 - 12(k+1)(k+9) = 0$$
Solve for $k$ and apply the $a \neq 0$ condition
Factor and solve, then reject any value that makes $a = 0$.
$$12(k+1)(2k - 6) = 0 \implies k = -1 \text{ or } k = 3$$
$$k = -1 \implies a = 0 \text{ (rejected)} \implies k = 3$$
14
If one root of $2x^2 + px - 8 = 0$ is $4$, find the other root and the value of $p$.
Ans
Other root $= -1$, $p = -6$
Review
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Find the other root using Vieta's product formula
Since $\alpha\beta = \frac{-8}{2} = -4$ and one root is $4$, solve for the other.
$$\beta = \frac{-4}{4} = -1$$
Find $p$ using Vieta's sum formula
The sum of roots $\alpha + \beta = -\frac{p}{a}$.
$$4 + (-1) = -\frac{p}{2} \implies 3 = -\frac{p}{2} \implies p = -6$$
15
A business's profit function is $P = -x^2 + 60x - 500$, where $x$ is the number of units produced. For what range of $x$ is the business profitable ($P > 0$)?
Ans
$10 < x < 50$
Review
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Set up the inequality
Rearrange $P > 0$ into standard quadratic form.
$$x^2 - 60x + 500 < 0$$
Find the roots and determine the interval
Factor to find where the quadratic changes sign, then test the region where it is negative.
$$(x - 10)(x - 50) < 0 \implies 10 < x < 50$$
16
For what values of $k$ does $x^2 - 2kx + k^2 - 1 = 0$ have real and distinct roots?
Ans
All real values of $k$
Review
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Compute the discriminant
Expand $b^2 - 4ac$ with $a = 1$, $b = -2k$, $c = k^2 - 1$.
$$\Delta = 4k^2 - 4(k^2 - 1) = 4$$
Interpret the result
Since $\Delta = 4 > 0$ for all real $k$, no restriction is needed.
$$\Delta = 4 > 0 \text{ for all } k \in \mathbb{R}$$
17
A factory's revenue and cost models (in thousands of rupees) are $R = x^2 + 10x$ and $C = 3x^2 - 2x + 30$, where $x$ is units produced. The break-even equation $R = C$ simplifies to $2x^2 - 12x + 30 = 0$. What is the discriminant, and can the factory break even?
Ans
$\Delta = -96$; the factory cannot break even
Review
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Compute the discriminant of the break-even equation
Apply $\Delta = b^2 - 4ac$ to $2x^2 - 12x + 30 = 0$.
$$\Delta = 144 - 240 = -96$$
Interpret for the business context
A negative discriminant means no real solution exists, so cost always exceeds revenue.
$$\Delta < 0 \implies \text{no break-even point exists}$$
18
A store owner finds that daily customer count varies with price as $C = 200 - 4p$ rupees. Daily revenue is $R = p \times C$. At what price (in rupees) does daily revenue equal Rs. 2400?
Ans
Rs. 20 or Rs. 30
Review
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Set up the revenue equation
Substitute the demand function into $R = p \times C$ and set equal to 2400.
$$p(200 - 4p) = 2400 \implies p^2 - 50p + 600 = 0$$
Solve for price
Factor the quadratic to find the two valid prices.
$$(p - 20)(p - 30) = 0 \implies p = 20 \text{ or } p = 30$$
19
For which integer values of $m$ does $x^2 + mx + m + 1 = 0$ have no real roots?
Ans
$m = 0, 1, 2, 3, 4$
Review
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Set up the discriminant inequality
Require $\Delta = b^2 - 4ac < 0$ for the equation to have no real roots.
$$m^2 - 4m - 4 < 0$$
Solve the inequality
Find the roots of the quadratic in $m$ and determine the interval where it is negative.
$$(m - (2 - 2\sqrt{2}))(m - (2 + 2\sqrt{2})) < 0$$
$$-0.83 < m < 4.83 \implies m = 0, 1, 2, 3, 4$$
20
If the roots of $x^2 - (m+2)x + (m+5) = 0$ are reciprocals of each other, find $m$.
Ans
$m = -4$
Review
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Use the reciprocal root condition
If roots are reciprocals, their product must equal 1.
$$\alpha\beta = 1$$
Apply Vieta's product formula
Set $\frac{c}{a} = m + 5$ equal to 1 and solve for $m$.
$$m + 5 = 1 \implies m = -4$$
Question 1 of 20
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1
Which of the following is a quadratic equation in standard form $ax^2 + bx + c = 0$ where $a \neq 0$?