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2. Solving Quadratic Equations by Completing the Square

For quadratic equations that cannot be solved by factorising, we use a method which can solve ALL quadratic equations called completing the square. We use this later when studying circles in plane analytic geometry.

Completing the square comes from considering the special formulas that we met in Square of a sum and square of a difference earlier:

(x + y)2 = x2 + 2xy + y2 (Square of a sum)

(xy)2 = x2 − 2xy + y2 (Square of a difference)

To find the roots of a quadratic equation in the form:

`ax^2+ bx + c = 0`,

follow these steps:

(i) If a does not equal `1`, divide each side by a (so that the coefficient of the x2 is `1`).

(ii) Rewrite the equation with the constant term on the right side.

(iii) Complete the square by adding the square of one-half of the coefficient of x to both sides.

(iv) Write the left side as a square and simplify the right side.

(v) Equate and solve.

Example 1

Find the roots of x2 + 10x − 4 = 0 using completing the square method.

Answer

Step (i) a = 1 [no action necessary in this example]

Step (ii) Rewrite the equation with the constant term on the right side.

x2 + 10x = 4

Step (iii) Complete the square by adding the square of one-half of the coefficient of x to both sides. In this case:

`(10/2)^2=5^2=25`

`x^2+ 10x + 25 = 4 + 25`

`x^2+ 10x + 25 = 29 `

Step (iv) Write the left side as a square:

`(x + 5)^2= 29`

Step (v) Equate and solve

`x+5=+-sqrt(29)`

`x=-5+-sqrt(29)`

Example 2

Solve 4x2 + x = 3 by completing the square.

Answer

Step (i) Divide each side by a which is 4 (so that the coefficient of the x2 is 1)

`x^2+x/4=3/4`

Step (ii) Rewrite the equation with the constant term (ie. 'c') on the right side.

[No need in this example]

Step (iii) Complete the square by adding the square of one-half of the coefficient of x to both sides, that is `(b/2)^2`.

`x^2+x/4+(1/8)^2=3/4+(1/8)^2`

`x^2+x/4+1/64=3/4+1/64`

Step (iv) Write the left side as a square and simplify the right side.

`(x+1/8)^2=(48+1)/64=49/64`

Step (v) Equate and solve

`x+1/8=+-sqrt(49/64)=+-7/8`

So

`x=-1/8+7/8=3/4`

or

`x=-1/8-7/8=-1`

This gives `x=3/4` or `x=-1`.

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Exercises

Solve the following quadratic equations by completing the square

Q1. `2s^2+ 5s = 3`

Answer

`2s^2+5s=3`

Divide throughout by 2.

`s^2+5/2s=3/2`

Take `1/2` of `5/2`, square it and add to both sides.

`s^2+5/2s+(5/4)^2=3/2+(5/4)^2`

Write the left side as a perfect square.

`(s+5/4)^2=3/2+25/16=(24+25)/16=49/16`

Solve for `s`.

`s+5/4=+-sqrt(49/16)`

`s=-5/4+-7/4=(-12)/4\ "or"\ 2/4`

`s=-3\ text(or)\ 1/2`

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Q2. `3x^2= 3 − 4x`

Answer

`3x^2=3-4x`

Rearrange:

`3x^2+4x=3`

Divide throughout by 3:

`x^2+4/3x=1`

Write left hand side as a perfect square:

`x^2+4/3x+(2/3)^2=1+(2/3)^2`

`(x+2/3)^2=1+4/9=13/9`

Solve:

`x+2/3=+-sqrt(13/9)=+-sqrt(13)/3`

`x=-2/3+-sqrt(13)/3`

`x=-1.869\ "or"\ x=0.535`

Q3. `9v^2− 6v − 2 = 0`

Answer

`9v^2-6v-2=0`

Rearrange:

`9v^2-6v=2`

Divide throughout by 9:

`v^2-2/3v=2/9`

Write as a perfect square:

`v^2-2/3v+(1/3)^2=2/9+(1/3)^2`

`(v-1/3)^2=2/9+1/9=1/3`

Solve:

`v-1/3 = +-sqrt(1/3)`

`v=1/3+-sqrt(1/3)`

`v=-0.244\ "or"\ v=0.911`

Q4. `ax^2+bx+c=0`

Answer

`ax^2+bx+c=0`

This is a general quadratic equation.

Rearrange:

`ax^2+bx=-c`

Divide throughout by `a`:

`x^2+b/a x =-c/a`

Write as a perfect square:

`x^2 + b/a x +(b/(2a))^2=-c/a+(b/(2a))^2`

`(x+b/(2a))^2=(-4ac+b^2)/(4a^2)`

Solve:

`x+b/(2a)= +-sqrt(-4ac+b^2)/(2a)`

`x=-b/(2a)+-sqrt(b^2-4ac)/(2a)`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

We'll use this result a great deal throughout the rest of the math we study.

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