Find one value of x that is a solution to the equation
(7x+2)^2+6(7x+2)=27
x= ?
2 answers:
Answer:
x = -11/7 and x = 1/7
Step-by-step 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
Answer:
x = 1/7 is one solution.
Step-by-step 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.
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