Solving ax^2+bx=0 and x^2=d
Theory
To solve \(x^2 = d\), take the square root of both sides: \(x = \pm\sqrt{d}\). To solve \(ax^2 + bx = 0\), take out the common factor \(x\) and use the null factor law: if two factors multiply to zero, one of them must be zero. This Year 9 maths page covers exact and rounded answers, squared brackets, area problems, why you must never divide by \(x\), and how many solutions an equation has.
A quadratic equation can be written as \(ax^2 + bx + c = 0\) with \(a \ne 0\). Its solutions (or roots) are the values of \(x\) that make it true.
When there is no \(x\) term, the equation can be written as \(x^2 = d\). If \(d > 0\) there are two solutions, \(x = \pm\sqrt{d}\); if \(d = 0\) there is one, \(x = 0\); if \(d \lt 0\) there are none, because a square is never negative. Give \(\pm\sqrt{d}\) exactly (a whole number, a fraction or a square root such as \(\sqrt{7}\)) or as a rounded decimal.
The null factor law: if \(A \times B = 0\), then \(A = 0\) or \(B = 0\). It only works when one side of the equation is zero.
When there is no constant term, as in \(ax^2 + bx = 0\), every term contains \(x\). Take out the common factor: \(x(ax + b) = 0\), so \(x = 0\) or \(ax + b = 0\). One solution is always \(x = 0\).
Never divide both sides by \(x\). \(x\) might be zero, and dividing by it throws that solution away.
The solutions of a quadratic equation are the \(x\)-intercepts of its parabola, so you can also read them from a graph.
| Equation | Rearranged | Solutions |
|---|---|---|
| \(x^2 - 64 = 0\) | \(x^2 = 64\) | \(x = \pm 8\) |
| \(x^2 + 7x = 0\) | \(x(x + 7) = 0\) | \(x = 0\) or \(-7\) |
| \(3x^2 - 15x = 0\) | \(3x(x - 5) = 0\) | \(x = 0\) or \(5\) |
| \((x - 1)^2 = 25\) | \(x - 1 = \pm 5\) | \(x = 6\) or \(-4\) |
How to solve \(x^2 = d\) and \(ax^2 = c\)
- Isolate \(x^2\) (divide by \(a\)).
- Look at the sign of the number equal to \(x^2\): positive gives two solutions, zero gives one, negative gives none.
- Take the square root of both sides and write \(\pm\). Leave the root exact, or round as asked.
How to solve \(ax^2 + bx = 0\)
- Make one side zero. Move every term to one side.
- Never divide by \(x\). You may divide by a number that every term shares.
- Take out the highest common factor, including \(x\): \(4x^2 + 10x = 2x(2x + 5)\).
- Set each factor to zero and solve each linear equation.
- Check both answers in the original equation.
Variations
- Squared bracket \((x + p)^2 = k\): get the bracket alone, take the square root first, then solve \(x + p = \sqrt{k}\) and \(x + p = -\sqrt{k}\). There is no need to expand.
- An area or a radius: write the formula, get the squared pronumeral on its own, then take the square root.
- A common bracket: bring everything to one side and factorise out the bracket.
- How many solutions? With a constant such as \(k\) in the equation, make \(x^2\) the subject and ask when the other side is positive, zero or negative.
- In context: reject a negative length or time, and say why.
(i) \(x^2 = 1.44\)
(ii) \(7x^2 = 63\)
(iii) \(9x^2 = 16\)
(iv) \(x^2 = 45\), exactly and correct to one decimal place
(i) \(1.2^2 = 1.44\), so
| \(x\) | \(=\) | \(\mathbf{\pm 1.2}\) |
(ii) Divide by \(7\) first.
| \(x^2\) | \(=\) | \(9\) |
| \(x\) | \(=\) | \(\mathbf{\pm 3}\) |
(iii) Divide by \(9\), then root the top and the bottom.
| \(x^2\) | \(=\) | \(\dfrac{16}{9}\) |
| \(x\) | \(=\) | \(\boldsymbol{\pm \dfrac{4}{3}}\) |
(iv) \(45\) is not a square number, so keep the root for the exact answer.
| \(x\) | \(=\) | \(\mathbf{\pm\sqrt{45}}\) |
| \(\) | \(\approx\) | \(\mathbf{\pm 6.7}\) |
(i) \(x^2 - 5x = 0\)
(ii) \(4x^2 + 10x = 0\)
(iii) \(16x^2 = 9x\)
(i) Take out the common factor \(x\).
| \(x(x - 5)\) | \(=\) | \(0\) |
| \(x = 0\) | \(\text{or}\) | \(x - 5 = 0\) |
| \(\boldsymbol{x = 0}\) | \(\text{or}\) | \(\boldsymbol{x = 5}\) |
(ii) The highest common factor is \(2x\).
| \(2x(2x + 5)\) | \(=\) | \(0\) |
| \(2x = 0\) | \(\text{or}\) | \(2x + 5 = 0\) |
| \(\boldsymbol{x = 0}\) | \(\text{or}\) | \(\boldsymbol{x = -\dfrac{5}{2}}\) |
(iii) Do not divide by \(x\): that would give only \(x = \dfrac{9}{16}\) and lose \(x = 0\). Make one side zero instead.
| \(16x^2 - 9x\) | \(=\) | \(0\) |
| \(x(16x - 9)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{0} \ \text{or}\ \boldsymbol{\dfrac{9}{16}}\) |
(i) \((x + 4)^2 = 49\)
(ii) \((x - 2)^2 - 25 = 0\)
(iii) A circle has an area of \(50~\text{cm}^2\). Find its radius, correct to \(1\) decimal place.
(i) Take the square root of both sides, keeping \(\pm\).
| \(x + 4\) | \(=\) | \(\pm 7\) |
| \(x = -4 + 7\) | \(\text{or}\) | \(x = -4 - 7\) |
| \(\boldsymbol{x = 3}\) | \(\text{or}\) | \(\boldsymbol{x = -11}\) |
(ii) Get the bracket on its own first.
| \((x - 2)^2\) | \(=\) | \(25\) |
| \(x - 2\) | \(=\) | \(\pm 5\) |
| \(x\) | \(=\) | \(\boldsymbol{7} \ \text{or}\ \boldsymbol{-3}\) |
(iii) Use \(A = \pi r^2\) and get \(r^2\) on its own. A radius is positive, so take only the positive root.
| \(\pi r^2\) | \(=\) | \(50\) |
| \(r^2\) | \(=\) | \(\dfrac{50}{\pi} = 15.915\ldots\) |
| \(r\) | \(=\) | \(\sqrt{15.915\ldots} = 3.989\ldots\) |
| \(r\) | \(\approx\) | \(\mathbf{4.0}~\textbf{cm}\) |
Check: \(\pi \times 3.989^2 \approx 50\ \checkmark\)
(i) Solve \((x + 1)^2 = 3x(x + 1)\).
(ii) For which values of \(k\) does \(2x^2 + k = 8\) have two solutions, one solution, or no solutions?
(i) Do not divide by \(x + 1\). Bring everything to one side and take out the common bracket.
| \((x + 1)^2 - 3x(x + 1)\) | \(=\) | \(0\) |
| \((x + 1)\big[(x + 1) - 3x\big]\) | \(=\) | \(0\) |
| \((x + 1)(1 - 2x)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-1} \ \text{or}\ \boldsymbol{\dfrac{1}{2}}\) |
Check \(x = \dfrac{1}{2}\): \(\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\) and \(3 \times \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{9}{4}\ \checkmark\)
(ii) Make \(x^2\) the subject.
| \(2x^2\) | \(=\) | \(8 - k\) |
| \(x^2\) | \(=\) | \(\dfrac{8 - k}{2}\) |
Two solutions when \(8 - k > 0\), that is \(k \lt 8\). One solution (\(x = 0\)) when \(k = 8\). No solutions when \(k > 8\).
Common pitfalls
Frequently asked questions
What is the null factor law?
If the product of two factors is zero, then at least one of the factors is zero. So if \((x - 2)(x + 5) = 0\), then \(x = 2\) or \(x = -5\).
Why can't I divide both sides by x?
Because \(x\) might be zero, and dividing by zero is not allowed. Dividing \(x^2 = 7x\) by \(x\) gives only \(x = 7\) and throws away \(x = 0\). Write \(x(x - 7) = 0\) instead.
How do you solve x squared = d when d is not a square number?
Take the square root of both sides and write \(\pm\). Give the answer exactly, as a square root, or as a rounded decimal. For \(x^2 = 20\), \(x = \pm\sqrt{20} \approx \pm 4.47\).
How do you solve (x + 2) squared = 9?
Take the square root of both sides: \(x + 2 = 3\) or \(x + 2 = -3\). So \(x = 1\) or \(x = -5\). There is no need to expand the bracket.
How many solutions can these equations have?
\(x^2 = d\) has two solutions when \(d\) is positive, one (\(x = 0\)) when \(d = 0\) and none when \(d\) is negative. \(ax^2 + bx = 0\) with \(b \ne 0\) always has two: \(x = 0\) and \(x = -\dfrac{b}{a}\).