# 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}=x^{2}+ 2xy+y^{2}(Square of a sum)(

x−y)^{2}=x^{2}− 2xy+y^{2}(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 *x*^{2}
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 *x*^{2} + 10*x* − 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.

x^{2}+ 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)`

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### Example 2

Solve 4*x*^{2} + *x* = 3 by completing the
square.

Answer

Step (i) Divide each side by *a* which is 4 (so that the coefficient of the *x*^{2} 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`

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**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`

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**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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