Quadratic Equations

A step-by-step journey — from zero to factoring

A When a product equals zero, solving is easy
Key idea: If $A \times B = 0$, then either $A = 0$ or $B = 0$ (or both). This is the zero-product property. Once an equation is written as a product, we just set each factor to zero.
A1 Solve   $x - 5 = 0$ Starter ▼

Find the value of $x$ that makes this true: $$x - 5 = 0$$

Hint: What number minus 5 gives zero?
Show solution
Add 5 to both sides: $x = 5$

Answer: $x = 5$

A2 Solve   $2(x - 3) = 0$ Starter ▼

$$2(x - 3) = 0$$

Hint: We have $A \times B = 0$ where $A = 2$ and $B = (x-3)$. Can $A = 2$ ever equal zero? What about $B$?
Show solution
$2 \neq 0$, so we need $x - 3 = 0$
$x = 3$

Answer: $x = 3$

A3 Solve   $(x - 3)(50 \cdot 10 + 10 - 5) = 0$ Easy ▼

$$(x - 3)\,(50 \cdot 10 + 10 - 5) = 0$$

Hint: Simplify the second factor first. Is it zero?
Show solution
Compute the second factor: $50 \times 10 + 10 - 5 = 500 + 10 - 5 = 505$
Equation becomes: $(x-3) \cdot 505 = 0$
$505 \neq 0$, so we need $x - 3 = 0$

Answer: $x = 3$

No matter how big or complicated a constant factor looks, if it isn't zero, only the variable factor matters.

A4 Solve   $(x - 1)(x - 2) = 0$ Easy ▼

$$(x-1)(x-2) = 0$$

Hint: Now both factors contain $x$. Either one could be zero. Try each case separately.
Show solution
Case 1: $x - 1 = 0 \;\Rightarrow\; x = 1$
Case 2: $x - 2 = 0 \;\Rightarrow\; x = 2$

Answer: $x = 1$   or   $x = 2$

A quadratic equation can have two solutions. Both are valid.

A5 Solve   $10\!\left(x - \tfrac{1}{2}\right)\!\left(x - \tfrac{2}{3}\right) = 0$ Medium ▼

$$10\left(x - \frac{1}{2}\right)\!\left(x - \frac{2}{3}\right) = 0$$

Hint: The constant factor 10 cannot be zero. Apply the zero-product property to the two remaining factors.
Show solution
$10 \neq 0$, so focus on the two brackets.
Case 1: $x - \dfrac{1}{2} = 0 \;\Rightarrow\; x = \dfrac{1}{2}$
Case 2: $x - \dfrac{2}{3} = 0 \;\Rightarrow\; x = \dfrac{2}{3}$

Answer: $x = \dfrac{1}{2}$   or   $x = \dfrac{2}{3}$

A6 Solve   $(x-4)(x-5)(x-6) = 0$ Medium ▼

$$(x-4)(x-5)(x-6) = 0$$

Hint: Three factors — any one of them could be zero. Find all three cases.
Show solution
Case 1: $x - 4 = 0 \;\Rightarrow\; x = 4$
Case 2: $x - 5 = 0 \;\Rightarrow\; x = 5$
Case 3: $x - 6 = 0 \;\Rightarrow\; x = 6$

Answer: $x = 4$,   $x = 5$,   or   $x = 6$

This is a cubic equation with three solutions — one per factor.

A7 Solve   $(x + 3)(x - 7) = 0$ Easy ▼

$$(x + 3)(x - 7) = 0$$

Hint: $(x+3)$ is the same as $(x - (-3))$. Set each factor to zero.
Show solution
Case 1: $x + 3 = 0 \;\Rightarrow\; x = -3$
Case 2: $x - 7 = 0 \;\Rightarrow\; x = 7$

Answer: $x = -3$   or   $x = 7$

One solution is negative, one is positive — both are perfectly valid.

A8 Solve   $(2x - 1)(3x + 6) = 0$ Medium ▼

$$(2x - 1)(3x + 6) = 0$$

Hint: Set each factor to zero. When you get $2x - 1 = 0$, solve for $x$ by adding 1 then dividing by 2.
Show solution
Case 1: $2x - 1 = 0 \;\Rightarrow\; 2x = 1 \;\Rightarrow\; x = \dfrac{1}{2}$
Case 2: $3x + 6 = 0 \;\Rightarrow\; 3x = -6 \;\Rightarrow\; x = -2$

Answer: $x = \dfrac{1}{2}$   or   $x = -2$

A10 Solve   $x(x - 1)(x + 5) = 0$ Medium ▼

$$x(x - 1)(x + 5) = 0$$

Hint: Notice the first factor is just $x$ — that is itself a factor! There are three factors total.
Show solution
Case 1: $x = 0$
Case 2: $x - 1 = 0 \;\Rightarrow\; x = 1$
Case 3: $x + 5 = 0 \;\Rightarrow\; x = -5$

Answer: $x = 0$,   $x = 1$,   or   $x = -5$

B Given $a+b$ and $ab$, can you find $a$ and $b$?
We will build three algebraic identities like stepping stones, then use them to recover $a$ and $b$ from their sum and product — a key skill for factoring quadratics.
B1 Expand   $(a+b)^2$ Starter ▼

Multiply out $(a+b)^2 = (a+b)(a+b)$. What do you get?

Hint: Use FOIL: First, Outer, Inner, Last.
Show solution
$(a+b)(a+b) = a^2 + ab + ba + b^2$
$= a^2 + 2ab + b^2$

$$(a+b)^2 = a^2 + 2ab + b^2$$

B2 Expand   $(a-b)^2$ Starter ▼

Now expand $(a-b)^2 = (a-b)(a-b)$. What do you get?

Hint: Same idea as B1, but watch the signs carefully.
Show solution
$(a-b)(a-b) = a^2 - ab - ba + b^2$
$= a^2 - 2ab + b^2$

$$(a-b)^2 = a^2 - 2ab + b^2$$

B3 Simplify   $(a+b)^2 - (a-b)^2$ Easy ▼

Using your answers from B1 and B2, compute: $$(a+b)^2 - (a-b)^2$$

Hint: Substitute the expanded forms and collect like terms.
Show solution
$(a^2 + 2ab + b^2) - (a^2 - 2ab + b^2)$
$= a^2 + 2ab + b^2 - a^2 + 2ab - b^2$
$= 4ab$

$$(a+b)^2 - (a-b)^2 = 4ab$$

Rearranging: $(a-b)^2 = (a+b)^2 - 4ab$

B4 Find $a$ and $b$ when $a+b=5,\; ab=6$ Medium ▼

You know two things about $a$ and $b$: $$a + b = 5 \qquad \text{and} \qquad ab = 6$$ Find $a$ and $b$ individually.

Hint: From B3 you know $(a-b)^2 = (a+b)^2 - 4ab$. Plug in the numbers to compute $(a-b)^2$, then find $a - b$. Once you have both $a+b$ and $a-b$, can you find $a$ and $b$? Try adding the two equations, then subtracting them.
Show solution
Step 1 — find $a - b$:
$(a-b)^2 = (a+b)^2 - 4ab = 5^2 - 4 \times 6 = 25 - 24 = 1$
$a - b = 1$
Step 2 — add the two equations:
$(a + b) + (a - b) = 5 + 1 = 6 = 2a \;\Rightarrow\; a = 3$
Step 3 — subtract the two equations:
$(a + b) - (a - b) = 5 - 1 = 4 = 2b \;\Rightarrow\; b = 2$

$a = 3,\; b = 2$

Check: $3 + 2 = 5$ ✓, $3 \times 2 = 6$ ✓

B5 Practice: find $a$ and $b$ when $a+b=7,\; ab=12$ Medium ▼

Given $a + b = 7$ and $ab = 12$, find $a$ and $b$.

Hint: Use the formula from B4.
Show solution
$(a-b)^2 = 7^2 - 4(12) = 49 - 48 = 1$
$a - b = 1$
$a = \dfrac{7+1}{2} = 4, \quad b = \dfrac{7-1}{2} = 3$

$a = 4,\; b = 3$

B6 Practice: find $a$ and $b$ when $a+b=6,\; ab=8$ Medium ▼

Given $a + b = 6$ and $ab = 8$, find $a$ and $b$.

Show solution
$(a-b)^2 = 36 - 32 = 4$
$a - b = 2$
$a = 4, \quad b = 2$

$a = 4,\; b = 2$

B7 Expand   $(a + b)(a - b)$ Easy ▼

Multiply out $(a + b)(a - b)$. What pattern do you notice?

Hint: Use FOIL. Pay close attention to the middle terms.
Show solution
$(a+b)(a-b) = a^2 - ab + ab - b^2$
$= a^2 - b^2$

$$(a+b)(a-b) = a^2 - b^2$$

This is the difference of two squares identity. The middle terms cancel perfectly. Read it backwards: $a^2 - b^2 = (a+b)(a-b)$ — very useful for factoring!

B8 Practice: find $a$ and $b$ when $a+b=10,\; ab=16$ Easy ▼

Given $a + b = 10$ and $ab = 16$, find $a$ and $b$.

Hint: $(a-b)^2 = (a+b)^2 - 4ab$. Compute it, then find $a$ and $b$.
Show solution
$(a-b)^2 = 100 - 64 = 36 \;\Rightarrow\; a - b = 6$
$a = \dfrac{10 + 6}{2} = 8, \quad b = \dfrac{10 - 6}{2} = 2$

$a = 8,\; b = 2$

Check: $8 + 2 = 10$ ✓, $8 \times 2 = 16$ ✓

B9 Practice: find $a$ and $b$ when $a+b=3,\; ab=-10$ Medium ▼

Given $a + b = 3$ and $ab = -10$, find $a$ and $b$.

Hint: $ab$ is negative, so one of $a$, $b$ must be negative. The formula still works — just be careful with signs inside the square root.
Show solution
$(a-b)^2 = 9 - 4(-10) = 9 + 40 = 49 \;\Rightarrow\; a - b = 7$
$a = \dfrac{3 + 7}{2} = 5, \quad b = \dfrac{3 - 7}{2} = -2$

$a = 5,\; b = -2$

Check: $5 + (-2) = 3$ ✓, $5 \times (-2) = -10$ ✓

B10 Challenge: find $a$ and $b$ when $a+b = 1,\; ab = -\frac{3}{4}$ Hard ▼

Given $a + b = 1$ and $ab = -\dfrac{3}{4}$, find $a$ and $b$.

Hint: $(a-b)^2 = 1^2 - 4\!\left(-\dfrac{3}{4}\right)$. Simplify carefully.
Show solution
$(a-b)^2 = 1 + 3 = 4 \;\Rightarrow\; a - b = 2$
$a = \dfrac{1 + 2}{2} = \dfrac{3}{2}, \quad b = \dfrac{1 - 2}{2} = -\dfrac{1}{2}$

$a = \dfrac{3}{2},\; b = -\dfrac{1}{2}$

Check: $\dfrac{3}{2} + \left(-\dfrac{1}{2}\right) = 1$ ✓, $\dfrac{3}{2} \times \left(-\dfrac{1}{2}\right) = -\dfrac{3}{4}$ ✓

B11 Practice: find $a$ and $b$ when $a+b=8,\; ab=15$ Easy ▼

Given $a + b = 8$ and $ab = 15$, find $a$ and $b$.

Hint: $(a-b)^2 = (a+b)^2 - 4ab$. Compute it, then find $a$ and $b$.
Show solution
$(a-b)^2 = 64 - 60 = 4 \;\Rightarrow\; a - b = 2$
$a = \dfrac{8 + 2}{2} = 5, \quad b = \dfrac{8 - 2}{2} = 3$

$a = 5,\; b = 3$

Check: $5 + 3 = 8$ ✓, $5 \times 3 = 15$ ✓

B12 Practice: find $a$ and $b$ when $a+b=-5,\; ab=6$ Medium ▼

Given $a + b = -5$ and $ab = 6$, find $a$ and $b$.

Hint: The sum is negative and the product is positive — so both $a$ and $b$ must be negative. The formula still works; just follow it carefully.
Show solution
$(a-b)^2 = (-5)^2 - 4(6) = 25 - 24 = 1 \;\Rightarrow\; a - b = 1$
$a = \dfrac{-5 + 1}{2} = -2, \quad b = \dfrac{-5 - 1}{2} = -3$

$a = -2,\; b = -3$

Check: $(-2) + (-3) = -5$ ✓, $(-2) \times (-3) = 6$ ✓

B13 Practice: find $a$ and $b$ when $a+b=2,\; ab=-8$ Medium ▼

Given $a + b = 2$ and $ab = -8$, find $a$ and $b$.

Hint: The product is negative, so one of $a$, $b$ is positive and the other is negative. The negative product makes $(a-b)^2$ larger than $(a+b)^2$.
Show solution
$(a-b)^2 = 4 - 4(-8) = 4 + 32 = 36 \;\Rightarrow\; a - b = 6$
$a = \dfrac{2 + 6}{2} = 4, \quad b = \dfrac{2 - 6}{2} = -2$

$a = 4,\; b = -2$

Check: $4 + (-2) = 2$ ✓, $4 \times (-2) = -8$ ✓

B14 Practice: find $a$ and $b$ when $a+b=0,\; ab=-9$ Medium ▼

Given $a + b = 0$ and $ab = -9$, find $a$ and $b$.

Hint: $a + b = 0$ means $b = -a$. You can verify your answer using the formula, or just substitute directly.
Show solution
$(a-b)^2 = 0 - 4(-9) = 36 \;\Rightarrow\; a - b = 6$
$a = \dfrac{0 + 6}{2} = 3, \quad b = \dfrac{0 - 6}{2} = -3$

$a = 3,\; b = -3$

When the sum is 0, the two numbers are always negatives of each other: $b = -a$. Check: $3 + (-3) = 0$ ✓, $3 \times (-3) = -9$ ✓

B15 Challenge: find $a$ and $b$ when $a+b=5,\; ab=\frac{9}{4}$ Hard ▼

Given $a + b = 5$ and $ab = \dfrac{9}{4}$, find $a$ and $b$.

Hint: $(a-b)^2 = 25 - 4 \cdot \dfrac{9}{4}$. Simplify carefully — the answer comes out cleanly.
Show solution
$(a-b)^2 = 25 - 4 \cdot \dfrac{9}{4} = 25 - 9 = 16 \;\Rightarrow\; a - b = 4$
$a = \dfrac{5 + 4}{2} = \dfrac{9}{2}, \quad b = \dfrac{5 - 4}{2} = \dfrac{1}{2}$

$a = \dfrac{9}{2},\; b = \dfrac{1}{2}$

Check: $\dfrac{9}{2} + \dfrac{1}{2} = 5$ ✓, $\dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{9}{4}$ ✓

C Converting a quadratic to factor form
The big idea: A quadratic $x^2 + Bx + C$ can always be written as $(x - a)(x - b)$. Expand the right side and match coefficients to find $a$ and $b$. Then use what we learned in Section B!
Key identity to remember
$$(x - a)(x - b) = x^2 \;-\; (a+b)\,x \;+\; ab$$ So if $x^2 + Bx + C = (x-a)(x-b)$, then: $$a + b = -B \qquad \text{and} \qquad ab = C$$
C1 Factor   $x^2 - 2x + 1$ Easy ▼

Write $x^2 - 2x + 1$ in the form $(x - a)(x - b)$.

Step-by-step guide:
1. Assume $x^2 - 2x + 1 = (x-a)(x-b)$.
2. Expand the right side.
3. Match coefficients to find $a + b$ and $ab$.
4. Use Section B to recover $a$ and $b$.
Show full solution
Expand:
$(x-a)(x-b) = x^2 - (a+b)x + ab$
Match with $x^2 - 2x + 1$:
$a + b = 2$    (coefficient of $x$ is $-2$, so $a+b = 2$)
$ab = 1$
Find $a - b$:
$(a-b)^2 = (a+b)^2 - 4ab = 4 - 4 = 0$
$a - b = 0$, so $a = b$
Recover $a$ and $b$:
$a = b = \dfrac{2}{2} = 1$

$$x^2 - 2x + 1 = (x-1)(x-1) = (x-1)^2$$

This is a perfect square. The equation $x^2 - 2x + 1 = 0$ has a single (repeated) solution $x = 1$.

C2 Factor   $x^2 - 5x + 6$ Medium ▼

$$x^2 - 5x + 6 = (x - a)(x - b)$$

Hint: What is $a+b$? What is $ab$? Find two numbers that add to 5 and multiply to 6.
Show solution
From the expansion: $a + b = 5$ and $ab = 6$
$(a-b)^2 = 25 - 24 = 1 \;\Rightarrow\; a - b = 1$
$a = 3,\; b = 2$

$$x^2 - 5x + 6 = (x-3)(x-2)$$

Solutions: $x = 3$ or $x = 2$

C3 Factor   $x^2 + 7x + 12$ Medium ▼

$$x^2 + 7x + 12 = (x - a)(x - b)$$

Hint: Here the $x$-coefficient is $+7$, so $-(a+b) = +7$, meaning $a+b = -7$. And $ab = 12$. Both $a$ and $b$ are negative.
Show solution
$a + b = -7$, $ab = 12$
$(a-b)^2 = 49 - 48 = 1 \;\Rightarrow\; a - b = \pm 1$
$a = -3,\; b = -4$ (or vice versa)

$$x^2 + 7x + 12 = (x+3)(x+4)$$

Solutions: $x = -3$ or $x = -4$

C4 Factor   $x^2 - x - 6$ Medium ▼

$$x^2 - x - 6 = (x - a)(x - b)$$

Hint: $a + b = 1$ and $ab = -6$. One of $a$, $b$ is negative!
Show solution
$a + b = 1$, $ab = -6$
$(a-b)^2 = 1 - 4(-6) = 1 + 24 = 25 \;\Rightarrow\; a-b = 5$
$a = 3,\; b = -2$

$$x^2 - x - 6 = (x-3)(x+2)$$

Solutions: $x = 3$ or $x = -2$

C5 Factor   $x^2 - 4x + 1$ Hard ▼

$$x^2 - 4x + 1 = (x - a)(x - b)$$

Hint: $a + b = 4$ and $ab = 1$. $(a-b)^2 = 16 - 4 = 12$. The roots are irrational — use $\sqrt{12} = 2\sqrt{3}$.
Show solution
$a + b = 4$, $ab = 1$
$(a - b)^2 = 16 - 4 = 12 \;\Rightarrow\; a - b = 2\sqrt{3}$
$a = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$, $\quad b = 2 - \sqrt{3}$

$$x^2 - 4x + 1 = \bigl(x - (2+\sqrt{3})\bigr)\bigl(x - (2-\sqrt{3})\bigr)$$

Solutions: $x = 2 \pm \sqrt{3}$

C6 General case: derive the quadratic formula Challenge ▼

Solve the general quadratic $$ax^2 + bx + c = 0, \quad a \neq 0$$ by dividing through by $a$ and applying the same method we used above.

Hint: Divide by $a$ to get $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$. Now set $S = -\dfrac{b}{a}$ and $P = \dfrac{c}{a}$, and use B4.
Show derivation
Divide by $a$: $\;x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$
From the key identity: roots satisfy $\alpha + \beta = -\dfrac{b}{a}$, $\;\alpha\beta = \dfrac{c}{a}$
$(\alpha - \beta)^2 = \left(\dfrac{b}{a}\right)^2 - \dfrac{4c}{a} = \dfrac{b^2 - 4ac}{a^2}$
$\alpha - \beta = \dfrac{\sqrt{b^2 - 4ac}}{a}$
$\alpha = \dfrac{-b/a + \sqrt{b^2-4ac}/a}{2} = \dfrac{-b + \sqrt{b^2-4ac}}{2a}$

$$\boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$$

This is the quadratic formula — and you just derived it from first principles!
The expression $b^2 - 4ac$ is called the discriminant. When it is negative, there are no real solutions.

C7 Factor   $x^2 - 16$ Easy ▼

$$x^2 - 16 = (x - a)(x - b)$$

Hint: $16 = 4^2$. This is a difference of two squares. Use the identity from B7.
Show solution
$a + b = 0$ and $ab = -16$
$(a - b)^2 = 0 - 4(-16) = 64 \;\Rightarrow\; a - b = 8$
$a = 4,\; b = -4$

$$x^2 - 16 = (x - 4)(x + 4)$$

Solutions: $x = 4$ or $x = -4$

Shortcut: whenever the $x$-term is missing and the constant is negative, look for difference of squares.

C8 Factor   $x^2 + 2x - 15$ Medium ▼

$$x^2 + 2x - 15 = (x - a)(x - b)$$

Hint: $a + b = -2$ and $ab = -15$. The product is negative, so $a$ and $b$ have opposite signs.
Show solution
$a + b = -2$, $ab = -15$
$(a-b)^2 = 4 - 4(-15) = 4 + 60 = 64 \;\Rightarrow\; a - b = 8$
$a = \dfrac{-2 + 8}{2} = 3, \quad b = \dfrac{-2 - 8}{2} = -5$

$$x^2 + 2x - 15 = (x - 3)(x + 5)$$

Solutions: $x = 3$ or $x = -5$

C9 Factor   $x^2 - 3x - \frac{10}{4}$ Hard ▼

$$x^2 - 3x - \frac{10}{4} = (x - a)(x - b)$$

Hint: $a + b = 3$ and $ab = -\dfrac{10}{4}$. You solved this exact pair of numbers in B10! (with different signs — check carefully.)
Show solution
$a + b = 3$, $\;ab = -\dfrac{10}{4} = -\dfrac{5}{2}$
$(a-b)^2 = 9 - 4\!\left(-\dfrac{5}{2}\right) = 9 + 10 = 19$
$a - b = \sqrt{19}$
$a = \dfrac{3 + \sqrt{19}}{2}, \quad b = \dfrac{3 - \sqrt{19}}{2}$

$$x^2 - 3x - \frac{5}{2} = \left(x - \frac{3+\sqrt{19}}{2}\right)\!\left(x - \frac{3-\sqrt{19}}{2}\right)$$

Solutions: $x = \dfrac{3 \pm \sqrt{19}}{2}$

C10 Factor   $2x^2 - 7x + 3$ Hard ▼

$$2x^2 - 7x + 3 = 0$$ The leading coefficient is not 1. Divide through by 2 first, then factor.

Hint: Divide by 2 to get $x^2 - \dfrac{7}{2}x + \dfrac{3}{2} = 0$. Now $a + b = \dfrac{7}{2}$ and $ab = \dfrac{3}{2}$.
Show solution
Divide by 2: $x^2 - \dfrac{7}{2}x + \dfrac{3}{2} = 0$
$a + b = \dfrac{7}{2}$, $\;ab = \dfrac{3}{2}$
$(a - b)^2 = \dfrac{49}{4} - 6 = \dfrac{49 - 24}{4} = \dfrac{25}{4}$
$a - b = \dfrac{5}{2}$
$a = \dfrac{7/2 + 5/2}{2} = \dfrac{6}{2} = 3, \quad b = \dfrac{7/2 - 5/2}{2} = \dfrac{1}{2}$

$$2x^2 - 7x + 3 = 2\!\left(x - 3\right)\!\left(x - \frac{1}{2}\right) = (x-3)(2x-1)$$

Solutions: $x = 3$ or $x = \dfrac{1}{2}$

C11 No real solutions: $x^2 + x + 1 = 0$ Challenge ▼

Try to factor $x^2 + x + 1$. What goes wrong? $$x^2 + x + 1 = (x - a)(x - b) \;?$$

Hint: Find $a + b$ and $ab$, then compute $(a - b)^2$. What sign does it have?
Show solution
$a + b = -1$, $\;ab = 1$
$(a - b)^2 = (-1)^2 - 4(1) = 1 - 4 = -3$
We need $(a-b)^2 = -3$, but a square can never be negative!

No real solutions exist.

The discriminant $b^2 - 4ac = 1 - 4 = -3 < 0$. When the discriminant is negative, the quadratic has no real roots — it cannot be factored over the real numbers. (In the complex numbers, the roots are $x = \dfrac{-1 \pm i\sqrt{3}}{2}$.)