Question

How do you solve `2x ^ { 2} + 10x + 52= ( x + 7) ^ { 2}`?

Answer 1

`x = 1, x =3`

Explanation:

`2x ^ { 2} + 10x + 52= ( x + 7) ^ { 2}`

FOIL the right side:

`2x^2+10x+52 = (x+7)(x+7)`

`2x^2+10x+52 = x^2 +14x +49`

Move the terms to one side of the equation:

`2x^2+10x+52 - x^2-14x -49= 0`

combine like terms:

`x^2 -4x +3 =0`

factor:

`(x-1)(x-3) = 0`

solutions (roots):

`x = 1, x =3`

Answer 2

"FOIL" terms on the right and subtract terms from both sides. Use quadratic formula to obtain answers.

Explanation:

`2x^2+10x+52=(x+7)^2`

`2x^2+10x+52=(x+7)(x+7)`

Foil the terms on the right (First, Outsides, Insides, Last).

`2x^2+10x+52=x^2+7x+7x+49`

`2x^2+10x+52=x^2+14x+49`

`x^2+10x+52=14x+49`

`x^2+52=4x+49`

`x^2+3=4x`

At this point, it may be obvious that one answer is `x=1`. The second answer may not be as obvious.

If not, we can always apply the Quadratic Formula, where `x^2-4x+3=0.`

`(4+sqrt(16-4*1*3))/(2*1)` or `(4-sqrt(16-4*1*3))/(2*1)`

`6/2=3` or `2/2=1.`

`x=3 or x=1`