Question

COMPLETING THE SQUARE

Answer 1

13.
`x=1,-7`

14.
`x=-3+√(14)` and
`x=-3-√(14)`

Explanation:

to solve the quadratic equation(
`ax^(2) +bx+c=0`) using completing the squares method:

Step1: divide the equation by a to make it in the form
`x^(2) +(b)/(a)x+(c)/(a) =0`

Step2: add
`((b)/(2a))^(2)` on both sides of the equation to get the eqaution:

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

Step3: rearrange them to get the square.

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

⇒
`x= -(b)/(2a)+\sqrt{((b)/(2a))^(2)-(c)/(a)}` and

`x= (b)/(2a)+\sqrt{((b)/(2a))^(2)-(c)/(a)}`

Now getting on to the question:

13.
`x^(2) +6x=7`

a=1; b=6; c=-7

adding
`(b)/(2a)^(2) = (6)/(2*1)^(2) =3^(2) =9` on both sides

⇒
`x^(2) +6x+9=7+9`

⇒
`x^(2) +6x+9=16`

⇒
`(x+3)^(2)=16`

⇒
`x+3=√(16)`

⇒
`x+3=4` and
`x+3=-4`

⇒
`x=1,-7`

14.
`x^(2) +6x=5`

a=1; b=6; c=-5

adding
`(b)/(2a)^(2) = (6)/(2*1)^(2) =3^(2) =9`on both sides

⇒
`x^(2) +6x+9=5+9`

⇒
`x^(2) +6x+9=14`

⇒
`(x+3)^(2)=14`

⇒
`x+3=√(14)`

⇒
`x+3=√(14)` and
`x+3=-√(14)`

⇒
`x=-3+√(14)` and
`x=-3-√(14)`