Question

Find one value of x that is a solution to the equation

(7x+2)^2+6(7x+2)=27

x= ?

Answer 1

x = 1/7 is one solution.

Explanation:

(7x+2)^2+6(7x+2)=27

Let 7x + 2 = y, then we have the quadratic equation:-

y^2 + 6y - 27 = 0

(y + 9)(y - 3) = 0

y = -9, 3

Substituting back for y:-

7x + 2 = -9

7x = -11

x = -11/7

Also 7x + 2 = 3

7x = 1

x = 1/7

Check if 1/7 fits the original equation:-

((7*-1/7 + 2)^2 + 6(7*1/7 + 2)

= 9 + 18

= 27 Checks OK.

Answer 2

x = -11/7 and x = 1/7

Explanation:

(7x+2)^2+6(7x+2)=27

Subtract 27 from each side

(7x+2)^2+6(7x+2)-27=0

Let m = 7x+2

m^2 +6m -27=0

Factor

(m+9) (m-3) = 0

Using the zero product property

m+9 = 0 m-3 =0

m=-9 m=3

Substitute m = 7x+2 back in

7x+2 = -9 7x+2 =3

Subtract 2 from each side

7x+2-2 = -9-2 7x+2-2 = 3-2

7x = -11 7x =1

Divide by 7

7x/7 = -11/7 7x/7 = 1/7

x = -11/7 and x = 1/7