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

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Answer 1 · 2018-05-22T00:31:35

#x = 1, x =3#

#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 · 2018-05-22T00:34:46

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

#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#