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Solving quadratic equations

Algebra · level 7, lesson 3 of 7

A question to start

Find a positive number that is exactly 1 more than its own reciprocal (the reciprocal of \(x\) is \(\dfrac1x\)).

Try 2: its reciprocal is 0.5, and 0.5 + 1 = 1.5, not 2. Try 1.5: \(\dfrac{1}{1.5} + 1 \approx 1.667\). Try 1.6: \(\dfrac{1}{1.6} + 1 = 1.625\). The answer is somewhere near 1.62, but guessing will never pin it down exactly. Written as an equation it is \(x = 1 + \dfrac1x\), and multiplying by \(x\) gives

\[x^2 = x + 1, \quad \text{or} \quad x^2 - x - 1 = 0.\]

It is a quadratic equation, and it does not factor with whole numbers. By the end of this lesson you will solve it exactly. The answer is a famous number, the golden ratio.

The idea

What it means to solve

A quadratic equation is one that can be written \(ax^2 + bx + c = 0\) with \(a \ne 0\). Its solutions are called its roots. On a graph, they are the zeros of the parabola \(y = ax^2 + bx + c\): the \(x\)-values where it meets the \(x\)-axis. A parabola can cross the axis twice, touch it once, or miss it, so a quadratic has two, one or no real roots.

There are three methods. All three start the same way: get 0 on one side.

Method 1: factoring and the zero product property

Zero product property: if \(A \times B = 0\), then \(A = 0\) or \(B = 0\) (or both).

Why? If \(A\) is not 0, divide both sides by \(A\): \(B = \dfrac{0}{A} = 0\). So one of them must be 0. This is special to zero. If \(A \times B = 6\), you learn nothing about either: it could be \(2 \times 3\) or \(12 \times 0.5\) or anything else.

So to solve \(x^2 - x - 12 = 0\), factor: \((x - 4)(x + 3) = 0\). Then \(x - 4 = 0\) or \(x + 3 = 0\), so \(x = 4\) or \(x = -3\).

Method 2: completing the square

When there are no nice factors, write the left side as a perfect square plus a number (lesson 2), then take square roots. Remember that a positive number has two square roots: if \(u^2 = 7\) then \(u = \sqrt7\) or \(u = -\sqrt7\), written \(u = \pm\sqrt7\).

\[x^2 + 4x - 3 = 0 \;\Rightarrow\; (x + 2)^2 - 4 - 3 = 0 \;\Rightarrow\; (x + 2)^2 = 7 \;\Rightarrow\; x = -2 \pm \sqrt7.\]

Method 3: the quadratic formula, derived

Do completing the square once, with letters, and you never have to do it again. Start with \(ax^2 + bx + c = 0\), \(a \ne 0\).

Divide every term by \(a\), and move the constant across:

\[x^2 + \frac{b}{a}x = -\frac{c}{a}.\]

Half of \(\dfrac{b}{a}\) is \(\dfrac{b}{2a}\). Add its square to both sides to complete the square:

\[x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}.\]

The left side is a perfect square. On the right, use the common denominator \(4a^2\):

\[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}.\]

Take square roots (both of them), using \(\sqrt{4a^2} = 2\lvert a \rvert\); the \(\pm\) already covers both signs:

\[x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}.\]

Subtract \(\dfrac{b}{2a}\):

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

This is the quadratic formula. It solves every quadratic equation, and now you know where it comes from.

The discriminant

The number under the square root, \(\Delta = b^2 - 4ac\), is called the discriminant (Δ is the Greek capital letter delta). It "discriminates", tells apart, the three cases:

x y Δ > 0: two roots x y Δ = 0: one root x y Δ < 0: no real roots
  • \(\Delta > 0\): \(\sqrt{\Delta}\) is a positive number, and \(+\sqrt\Delta\) and \(-\sqrt\Delta\) give two different roots. The parabola crosses the axis twice.
  • \(\Delta = 0\): both choices give \(x = -\dfrac{b}{2a}\), one repeated root. The parabola just touches the axis at its vertex.
  • \(\Delta < 0\): no real number squares to a negative, so there are no real roots. The parabola misses the axis.

The formula also shows that the two roots sit equally far either side of \(x = -\dfrac{b}{2a}\): that is the axis of symmetry, and the vertex formula promised in lesson 2.

Which method to choose

  • If it factors easily, factor: it is fastest and gives exact answers.
  • If \(b\) is even and \(a = 1\), completing the square is quick.
  • Otherwise, use the formula. It always works.

Worked examples

1. Factoring. Solve \(x^2 = x + 12\).

  • Get 0 on one side: \(x^2 - x - 12 = 0\).
  • Two numbers with product \(-12\) and sum \(-1\): \(-4\) and \(3\). So \((x - 4)(x + 3) = 0\).
  • \(x = 4\) or \(x = -3\). Check \(x = -3\): \(9 = -3 + 12\). ✓

2. Factoring with \(a \ne 1\). Solve \(6x^2 + x - 2 = 0\).

  • \(ac = 6 \times (-2) = -12\). Two numbers with product \(-12\) and sum \(1\): \(4\) and \(-3\).
  • Split the middle: \(6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2)\).
  • \(x = \dfrac12\) or \(x = -\dfrac23\).

3. Completing the square. Solve \(x^2 + 4x - 3 = 0\).

  • \((x + 2)^2 = 7\), so \(x = -2 \pm \sqrt7\), that is \(x \approx 0.646\) or \(x \approx -4.646\).

4. The formula. Solve \(2x^2 - 5x - 4 = 0\).

  • \(a = 2\), \(b = -5\), \(c = -4\). \(\Delta = (-5)^2 - 4(2)(-4) = 25 + 32 = 57\).
  • \(x = \dfrac{5 \pm \sqrt{57}}{4}\), so \(x \approx 3.137\) or \(x \approx -0.637\).

5. The golden ratio. Solve \(x^2 - x - 1 = 0\).

  • \(a = 1\), \(b = -1\), \(c = -1\). \(\Delta = 1 + 4 = 5\).
  • \(x = \dfrac{1 \pm \sqrt5}{2}\). The positive root is \(\dfrac{1 + \sqrt5}{2} \approx 1.618\), the golden ratio, usually called \(\varphi\) (phi). Check: \(\dfrac{1}{1.618} \approx 0.618\), and \(0.618 + 1 = 1.618\). ✓

6. A rectangle. A rectangle has area 60 cm², and its length is 7 cm more than its width. Find the width.

  • Width \(w\), length \(w + 7\): \(w(w + 7) = 60\), so \(w^2 + 7w - 60 = 0\), so \((w + 12)(w - 5) = 0\).
  • \(w = 5\) or \(w = -12\). A width cannot be negative, so the width is 5 cm (and the length 12 cm).

7. Using the discriminant. For what value of \(k\) does \(kx^2 + 4x + 1 = 0\) have exactly one root?

  • One root when \(\Delta = 0\): \(16 - 4k = 0\), so \(k = 4\). Then \(4x^2 + 4x + 1 = (2x + 1)^2\), root \(x = -\tfrac12\).

Try it

  1. Solve \(x^2 + 7x + 10 = 0\).
  2. Solve \(x^2 = 5x\).
  3. Solve \(3x^2 - 10x - 8 = 0\) by factoring.
  4. Solve \(x^2 - 6x + 4 = 0\) by completing the square. Give exact answers and decimals.
  5. Solve \(2x^2 + 3x - 7 = 0\) with the formula, to 3 decimal places.
  6. Without solving, how many real roots has each? (a) \(x^2 + 2x + 5 = 0\) (b) \(4x^2 - 12x + 9 = 0\) (c) \(x^2 - 3x - 1 = 0\)
  7. A stone's height is \(h = -5t^2 + 20t + 25\) metres. When does it hit the ground (\(h = 0\))?
  8. For which values of \(k\) does \(x^2 + kx + 9 = 0\) have two different real roots?
Answers
  1. \((x + 2)(x + 5) = 0\), so \(x = -2\) or \(x = -5\).
  2. \(x^2 - 5x = 0\), so \(x(x - 5) = 0\), so \(x = 0\) or \(x = 5\). (Dividing both sides by \(x\) would lose the root \(x = 0\).)
  3. \(ac = -24\); the numbers \(-12\) and \(2\) have product \(-24\) and sum \(-10\). \(3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)\). So \(x = 4\) or \(x = -\dfrac23\).
  4. \((x - 3)^2 - 9 + 4 = 0\), so \((x - 3)^2 = 5\), so \(x = 3 \pm \sqrt5\): \(x \approx 5.236\) or \(x \approx 0.764\).
  5. \(\Delta = 9 - 4(2)(-7) = 9 + 56 = 65\). \(x = \dfrac{-3 \pm \sqrt{65}}{4}\): \(x \approx 1.266\) or \(x \approx -2.766\).
  6. (a) \(\Delta = 4 - 20 = -16 < 0\): none. (b) \(\Delta = 144 - 144 = 0\): one, \(x = \tfrac32\) (it is \((2x - 3)^2 = 0\)). (c) \(\Delta = 9 + 4 = 13 > 0\): two.
  7. Divide by \(-5\): \(t^2 - 4t - 5 = 0\), so \((t - 5)(t + 1) = 0\). \(t = 5\) or \(t = -1\); time after the throw is positive, so it lands after 5 seconds.
  8. Two roots when \(\Delta = k^2 - 36 > 0\), so \(k^2 > 36\), which means \(k > 6\) or \(k < -6\).

Common mistake

Using the zero product property when the right side is not 0. From \((x - 2)(x - 3) = 6\) you can not say \(x - 2 = 6\) or \(x - 3 = 6\). Expand and bring everything to one side first: \(x^2 - 5x + 6 = 6\), so \(x^2 - 5x = 0\), so \(x(x - 5) = 0\), giving \(x = 0\) or \(x = 5\).

Dividing by the variable. From \(x^2 = 5x\), dividing by \(x\) gives only \(x = 5\) and loses \(x = 0\). You may only divide by something you know is not zero.

Dividing only part of the formula by \(2a\). The whole of \(-b \pm \sqrt{\Delta}\) is over \(2a\), not just the square root.

Where it leads