Lesson 27 of 46 · Algebra

Quadratic equations: factoring, the formula and the discriminant

Solve quadratics by factoring or by the formula, keep both roots, read the discriminant, and use the sum and product of roots as a shortcut.

About 22 minutes

In this lesson

  • Rearrange a quadratic to equal zero and solve it by factoring and the zero-product rule.
  • Use the quadratic formula when factoring fails, and the discriminant to count real solutions.
  • Find the sum and product of the roots straight from the coefficients.
  • Keep every root, and use conditions in the stem to choose between them.

The idea

What makes an equation quadratic

A quadratic equation can be written as ax2+bx+c=0ax^2 + bx + c = 0 with a≠0a \neq 0. It can have two real solutions, one, or none. That is the key difference from a linear equation: there is usually more than one value of xx that works, and many questions turn on the second one.

Step one: make one side zero

Every method starts from ax2+bx+c=0ax^2 + bx + c = 0. Move all terms to one side first. The equation x2=5x+14x^2 = 5x + 14 becomes x2−5x−14=0x^2 - 5x - 14 = 0. Do not try to factor while terms are on both sides.

Factoring and the zero-product rule

If a product equals zero, at least one factor equals zero. So once a quadratic is factored, each bracket gives a root:

x2−5x−14=(x−7)(x+2)=0⇒x=7 or x=−2x^2 - 5x - 14 = (x - 7)(x + 2) = 0 \quad\Rightarrow\quad x = 7 \text{ or } x = -2

Watch the signs: the factor (x+2)(x + 2) gives the root −2-2, not 22. Then substitute each root back into the original equation; both should work.

The rule needs zero on one side. (x−1)(x−4)=10(x - 1)(x - 4) = 10 does not mean x−1=10x - 1 = 10 or x−4=10x - 4 = 10. Expand, move the 10 across and factor again: x2−5x−6=0x^2 - 5x - 6 = 0, so (x−6)(x+1)=0(x - 6)(x + 1) = 0 and x=6x = 6 or x=−1x = -1.

Never divide by the variable

Faced with 3x2=12x3x^2 = 12x, it is tempting to divide both sides by xx and write 3x=123x = 12. That loses a root. Dividing by xx is only allowed if x≠0x \neq 0, and here x=0x = 0 is a solution. Instead, move everything to one side and factor: 3x2−12x=3x(x−4)=03x^2 - 12x = 3x(x - 4) = 0, so x=0x = 0 or x=4x = 4.

Square roots give two signs

If a square equals a positive number, the thing being squared can be positive or negative. From (x−3)2=25(x - 3)^2 = 25, x−3=5x - 3 = 5 or x−3=−5x - 3 = -5, so x=8x = 8 or x=−2x = -2. Writing only x−3=5x - 3 = 5 is one of the most common ways to lose a root.

The quadratic formula

When the numbers do not factor neatly, use

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Write down aa, bb and cc with their signs before substituting, and put negative values in brackets. The whole numerator, −b-b and the root, is divided by 2a2a.

The discriminant counts the roots

The expression under the root, b2−4acb^2 - 4ac, is the discriminant.

b2−4acb^2 - 4ac Real solutions
positive two different ones
zero exactly one (a repeated root)
negative none

You can answer "how many solutions" questions without solving at all. For x2+4x+7=0x^2 + 4x + 7 = 0, the discriminant is 16−28=−1216 - 28 = -12, so there are no real solutions.

Sum and product of the roots

For ax2+bx+c=0ax^2 + bx + c = 0 with roots rr and ss:

r+s=−bars=car + s = -\frac{b}{a} \qquad rs = \frac{c}{a}

This follows from writing the quadratic as a(x−r)(x−s)a(x - r)(x - s) and expanding. It answers "what is the sum of the solutions" in one line, and it is a quick check on roots you found by factoring: for x2−5x−14=0x^2 - 5x - 14 = 0 the roots 77 and −2-2 add to 55 and multiply to −14-14, as they should. Check that the discriminant is not negative before you rely on it for real roots.

Two roots in a comparison

When a quantitative comparison asks about a variable defined by a quadratic, find both roots and compare each with the other quantity. If one root makes Quantity A greater and the other makes Quantity B greater, the relationship cannot be determined. If the stem adds a condition such as x>0x > 0, use it to discard a root before comparing.

Worked examples

Example 1: Rearrange, then factor

Solve x2=5x+14x^2 = 5x + 14.

  1. Make one side zero: x2−5x−14=0x^2 - 5x - 14 = 0.

  2. Find two numbers that multiply to −14-14 and add to −5-5: they are −7-7 and 22. So (x−7)(x+2)=0(x - 7)(x + 2) = 0.

  3. Zero-product rule: x=7x = 7 or x=−2x = -2.

  4. Check in the original: 49=35+1449 = 35 + 14 and 4=−10+144 = -10 + 14. Both hold. As a second check, the roots add to 5=−−515 = -\frac{-5}{1} and multiply to −14-14.

Answer

x=7x = 7 or x=−2x = -2. Reading the roots straight off the brackets as −7-7 and 22 is the usual sign slip.

Example 2: Do not divide by xx

Solve 3x2=12x3x^2 = 12x.

  1. Move every term to one side: 3x2−12x=03x^2 - 12x = 0.

  2. Take out the common factor: 3x(x−4)=03x(x - 4) = 0.

  3. So 3x=03x = 0 or x−4=0x - 4 = 0, giving x=0x = 0 or x=4x = 4.

  4. Check: x=0x = 0 gives 0=00 = 0, and x=4x = 4 gives 48=4848 = 48.

Answer

x=0x = 0 or x=4x = 4. Dividing both sides by xx gives only x=4x = 4 and silently loses the root 0.

Example 3: When factoring fails

Solve 2x2−4x−3=02x^2 - 4x - 3 = 0.

  1. No pair of integers multiplies to 2×(−3)=−62 \times (-3) = -6 and adds to −4-4, so use the formula with a=2a = 2, b=−4b = -4, c=−3c = -3.

  2. Discriminant: b2−4ac=16−4(2)(−3)=16+24=40b^2 - 4ac = 16 - 4(2)(-3) = 16 + 24 = 40. It is positive, so there are two real roots.

  3. x=4±404=4±2104=2±102x = \frac{4 \pm \sqrt{40}}{4} = \frac{4 \pm 2\sqrt{10}}{4} = \frac{2 \pm \sqrt{10}}{2}.

  4. Check with sum and product: the roots add to 42=2\frac{4}{2} = 2, which is −ba=42-\frac{b}{a} = \frac{4}{2}; they multiply to 4−104=−32\frac{4 - 10}{4} = -\frac{3}{2}, which is ca\frac{c}{a}.

Answer

x=2+102x = \frac{2 + \sqrt{10}}{2} or x=2−102x = \frac{2 - \sqrt{10}}{2}, roughly 2.582.58 and −0.58-0.58.

Example 4: A square equal to a number

If (x−3)2=25(x - 3)^2 = 25, what is the sum of all possible values of xx?

  1. Take square roots, keeping both signs: x−3=5x - 3 = 5 or x−3=−5x - 3 = -5.

  2. So x=8x = 8 or x=−2x = -2.

  3. Second route: expand to x2−6x+9=25x^2 - 6x + 9 = 25, so x2−6x−16=0x^2 - 6x - 16 = 0, whose roots add to −−61=6-\frac{-6}{1} = 6.

Answer

The sum is 8+(−2)=68 + (-2) = 6. Keeping only the positive square root gives 8, the tempting wrong answer.

Common traps

  • Dividing by the variable

    Dividing x2=7xx^2 = 7x by xx leaves x=7x = 7 and loses x=0x = 0. Never divide by an expression that could be zero; factor it out instead.

  • Only the positive square root

    x2=36x^2 = 36 means x=6x = 6 or x=−6x = -6. The symbol 36\sqrt{36} means only 66, but solving x2=36x^2 = 36 gives both. Comparisons often hinge on the negative root.

  • Zero-product rule without zero

    (x−2)(x+1)=4(x - 2)(x + 1) = 4 does not give x−2=4x - 2 = 4. The rule works only when the product equals zero. Expand, move everything to one side, then factor.

  • Sign of the roots read from the brackets

    The factor (x+5)(x + 5) gives the root −5-5. A root found with the wrong sign is often among the options, so substitute it back before choosing.

Check yourself

Answer each question before you open its answer.

Question 1: Choose the root the stem allows

Algebra

If x2+2x−35=0x^2 + 2x - 35 = 0 and x>0x > 0, what is the value of xx?

  1. A.−7-7
  2. B.−5-5
  3. C.55
  4. D.77
  5. E.3535
Show the answer

Answer: C. 55

Working

Two numbers that multiply to −35-35 and add to 22 are 77 and −5-5, so (x+7)(x−5)=0(x + 7)(x - 5) = 0 and x=−7x = -7 or x=5x = 5. Only 55 is positive. Check: 25+10−35=025 + 10 - 35 = 0.

Why the other options are wrong

  • A. This is a root, but the condition x>0x > 0 rules it out.
  • B. Reads the roots off (x+7)(x−5)(x + 7)(x - 5) with the signs reversed.
  • D. Reverses the sign of the root −7-7 from the factor (x+7)(x + 7); check: 49+14−35=2849 + 14 - 35 = 28, not 0.
  • E. This is the size of the constant term, not a root.

What carries over

Find both roots, then let the condition in the stem choose.

Question 2: A root you might lose

Quantitative Comparison

x2=6xx^2 = 6x

Quantity A Quantity B
xx 33
  1. A.Quantity A is greater.
  2. B.Quantity B is greater.
  3. C.The two quantities are equal.
  4. D.The relationship cannot be determined from the information given.
Show the answer

Answer: D. The relationship cannot be determined from the information given.

Working

x2−6x=0x^2 - 6x = 0, so x(x−6)=0x(x - 6) = 0 and x=0x = 0 or x=6x = 6. With x=0x = 0, Quantity B is greater; with x=6x = 6, Quantity A is greater. The relationship is not fixed.

Why the other options are wrong

  • A. Comes from dividing both sides by xx, which leaves only x=6x = 6 and loses the root x=0x = 0.
  • B. Holds for the root x=0x = 0 only; the root x=6x = 6 makes Quantity A greater.
  • C. x=3x = 3 is halfway between the roots, but it is not a solution: 9≠189 \neq 18.

What carries over

In a comparison, test every root of a quadratic; two roots on opposite sides of the other quantity mean it cannot be determined.

Question 3: Sum without solving

Algebra

What is the sum of the solutions of 2x2−10x+7=02x^2 - 10x + 7 = 0?

  1. A.−5-5
  2. B.72\frac{7}{2}
  3. C.55
  4. D.77
  5. E.1010
Show the answer

Answer: C. 55

Working

The discriminant is 100−56=44>0100 - 56 = 44 > 0, so there are two real roots, and their sum is −ba=−−102=5-\frac{b}{a} = -\frac{-10}{2} = 5. Check with the formula: the roots are 10±444=5±112\frac{10 \pm \sqrt{44}}{4} = \frac{5 \pm \sqrt{11}}{2}, which add to 102=5\frac{10}{2} = 5.

Why the other options are wrong

  • A. Uses ba\frac{b}{a} instead of −ba-\frac{b}{a}; the sum of the roots carries the opposite sign of bb.
  • B. This is ca\frac{c}{a}, the product of the roots, not their sum.
  • D. This is the constant term, which relates to the product of the roots, and even then it must be divided by a=2a = 2.
  • E. Takes −b-b without dividing by a=2a = 2.

What carries over

For ax2+bx+c=0ax^2 + bx + c = 0, the roots add to −ba-\frac{b}{a} and multiply to ca\frac{c}{a}.

Key points

  • Make one side zero before factoring; the zero-product rule needs a zero.

  • Never divide by the variable: factor it out and keep the root 0.

  • A square equal to a positive number gives two roots, one of each sign.

  • The discriminant b2−4acb^2 - 4ac tells you whether there are two, one or no real roots.

  • Roots add to −ba-\frac{b}{a} and multiply to ca\frac{c}{a}.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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