Solving Quadratics by Completing the Square
Solving by completing the square adds the square of half the coefficient of \(x\) to both sides, so the left side becomes a perfect square, then takes the square root of both sides. For \(x^2+6x-7=0\) this gives \((x+3)^2=16\), so \(x=1\) or \(x=-7\). This NSW Year 10 Advanced maths method also proves the quadratic formula (NESA outcome MA5‑EQU‑P‑02).
You will complete the square for equations with whole-number, fraction and surd answers, divide by the leading coefficient first, and rearrange equations with brackets and fractions. You will show that an equation has no real solutions, find when an equation with a constant has real solutions, find the width of a swimming pool, and solve equations with letters as coefficients. Finally you will derive the quadratic formula. It is part of Year 10 Advanced Quadratics.
Every question with a fully worked solution.
- Solving Quadratics by Completing the Square - Video - Solving Quadratic Equations By Completing The Square Watch
Theory
To solve a quadratic by completing the square, rewrite it as \((x+h)^2=k\), then take the square root of both sides: \(x=-h\pm\sqrt{k}\). This Year 10 Advanced maths page covers even and odd coefficients, non-monic quadratics, exact surd answers, equations with no real solutions and worded problems.
A perfect square is an expression such as \(x^2+10x+25=(x+5)^2\). The constant \(25\) is the square of half the coefficient of \(x\): half of \(10\) is \(5\), and \(5^2=25\).
Completing the square means adding that number to make a perfect square. In an equation, add it to both sides so the equation stays balanced.
Once the equation is \((x+h)^2=k\), take the square root of both sides. Both the positive and the negative root work, so \(x+h=\pm\sqrt{k}\).
If \(k\gt 0\) there are two solutions, if \(k=0\) there is one, and if \(k\lt 0\) there are no real solutions, because the square of a real number is never negative. The method only works directly when the coefficient of \(x^2\) is \(1\), so divide through by it first.
The key identity: add the square of half the coefficient of \(x\).
| Expression | Half of \(b\) | Add | Perfect square |
|---|---|---|---|
| \(x^2+12x\) | \(6\) | \(36\) | \((x+6)^2\) |
| \(x^2-14x\) | \(-7\) | \(49\) | \((x-7)^2\) |
| \(x^2+7x\) | \(\dfrac{7}{2}\) | \(\dfrac{49}{4}\) | \(\left(x+\dfrac{7}{2}\right)^2\) |
| \(x^2-\dfrac{2}{3}x\) | \(-\dfrac{1}{3}\) | \(\dfrac{1}{9}\) | \(\left(x-\dfrac{1}{3}\right)^2\) |
| \(x^2+6mx\) | \(3m\) | \(9m^2\) | \((x+3m)^2\) |
How to solve by completing the square
- Rearrange into \(ax^2+bx+c=0\): expand brackets and clear fractions.
- Divide every term by \(a\) if \(a\neq 1\).
- Move the constant to the right side: \(x^2+bx=-c\).
- Add \(\left(\dfrac{b}{2}\right)^2\) to both sides.
- Write the left side as \(\left(x+\dfrac{b}{2}\right)^2\) and simplify the right side to a single number \(k\).
- Check the sign of \(k\). If \(k\lt 0\) there are no real solutions. Otherwise take \(\pm\sqrt{k}\) and solve for \(x\). Simplify the surd.
Other question types
- Show no real solutions: complete the square and show that a square equals a negative number.
- Find the values of \(k\): write the quadratic side as \((x+h)^2+c\). Real solutions need the other side minus \(c\) to be at least \(0\) (Example 3).
- General results: the same steps work with letters. Applied to \(ax^2+bx+c=0\), dividing by \(a\) and adding \(\left(\dfrac{b}{2a}\right)^2\) is how the quadratic formula is built.
- Word problems: form the equation, complete the square, then reject any solution the situation does not allow (Example 4).
(i) \(x^2+6x-16=0\)
(ii) \(x^2-8x+3=0\)
(i) Move the constant. Half of \(6\) is \(3\), so add \(3^2=9\) to both sides.
| \(x^2+6x\) | \(=\) | \(16\) |
| \(x^2+6x+9\) | \(=\) | \(16+9\) |
| \((x+3)^2\) | \(=\) | \(25\) |
| \(x+3\) | \(=\) | \(\pm 5\) |
| \(x\) | \(=\) | \(\boldsymbol{2} \text{ or } \boldsymbol{x=-8}\) |
(ii) Half of \(-8\) is \(-4\), so add \((-4)^2=16\).
| \(x^2-8x\) | \(=\) | \(-3\) |
| \(x^2-8x+16\) | \(=\) | \(-3+16\) |
| \((x-4)^2\) | \(=\) | \(13\) |
| \(x-4\) | \(=\) | \(\pm\sqrt{13}\) |
| \(x\) | \(=\) | \(\boldsymbol{4\pm\sqrt{13}}\) |
(i) \(x^2+5x-3=0\)
(ii) \(2x^2+6x-3=0\)
(i) Half of \(5\) is \(\dfrac{5}{2}\), so add \(\dfrac{25}{4}\). Write \(3\) as \(\dfrac{12}{4}\).
| \(x^2+5x+\dfrac{25}{4}\) | \(=\) | \(\dfrac{12}{4}+\dfrac{25}{4}\) |
| \(\left(x+\dfrac{5}{2}\right)^2\) | \(=\) | \(\dfrac{37}{4}\) |
| \(x+\dfrac{5}{2}\) | \(=\) | \(\pm\dfrac{\sqrt{37}}{2}\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{-5\pm\sqrt{37}}{2}}\) |
(ii) First divide every term by \(2\), so the coefficient of \(x^2\) is \(1\).
| \(x^2+3x\) | \(=\) | \(\dfrac{3}{2}\) |
| \(x^2+3x+\dfrac{9}{4}\) | \(=\) | \(\dfrac{6}{4}+\dfrac{9}{4}\) |
| \(\left(x+\dfrac{3}{2}\right)^2\) | \(=\) | \(\dfrac{15}{4}\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{-3\pm\sqrt{15}}{2}}\) |
(i) Show that \(x^2-4x+7=0\) has no real solutions.
(ii) For which values of \(k\) does \(x^2+8x+20=k\) have real solutions? Solve it when \(k=13\).
(i) Complete the square.
| \(x^2-4x+4\) | \(=\) | \(-7+4\) |
| \((x-2)^2\) | \(=\) | \(-3\) |
A square is never negative, so \((x-2)^2=-3\) is impossible. There are no real solutions.
(ii) Write the left side in the form \((x+h)^2+c\): \(x^2+8x+20=(x+4)^2+4\).
| \((x+4)^2+4\) | \(=\) | \(k\) |
| \((x+4)^2\) | \(=\) | \(k-4\) |
The left side is at least \(0\), so real solutions need \(k-4\ge 0\), that is \(\boldsymbol{k\ge 4}\). When \(k=13\):
| \((x+4)^2\) | \(=\) | \(9\) |
| \(x+4\) | \(=\) | \(\pm 3\) |
| \(x\) | \(=\) | \(\boldsymbol{-1} \text{ or } \boldsymbol{x=-7}\) |
Area \(=\) width \(\times\) length:
| \(x(x+6)\) | \(=\) | \(30\) |
| \(x^2+6x+9\) | \(=\) | \(30+9\) |
| \((x+3)^2\) | \(=\) | \(39\) |
| \(x\) | \(=\) | \(-3\pm\sqrt{39}\) |
\(-3-\sqrt{39}\) is negative, but a width must be positive. Reject it.
| \(x\) | \(=\) | \(\boldsymbol{-3+\sqrt{39}}\) |
| \(\) | \(\approx\) | \(\boldsymbol{3.24}\text{ m}\) |
Length \(=x+6=3+\sqrt{39}\approx\boldsymbol{9.24}\) m. Check: \(3.245\times 9.245\approx 30\).
Common pitfalls
Frequently asked questions
What does completing the square mean?
It means adding the number that turns \(x^2+bx\) into a perfect square. That number is \(\left(\dfrac{b}{2}\right)^2\), so \(x^2+6x\) becomes \(x^2+6x+9=(x+3)^2\).
How do you complete the square when the coefficient of x is odd?
Use a fraction. Half of \(5\) is \(\dfrac{5}{2}\), so add \(\dfrac{25}{4}\). Write the other side over \(4\) as well so the fractions are easy to add.
What if the coefficient of x squared is not 1?
Divide every term of the equation by that coefficient first, then complete the square as usual. The fractions this creates are normal.
How does completing the square show there are no solutions?
If you reach a square equal to a negative number, such as \((x-2)^2=-3\), there is no real solution, because squaring a real number never gives a negative result.
Is completing the square related to the quadratic formula?
Yes. Completing the square on the general equation \(ax^2+bx+c=0\) produces the quadratic formula, so both methods always give the same solutions.