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

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

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Answer 1 · 2018-05-22T00:31:35

$x = 1, x = 3$ $x = 1, x =3$

Explanation:

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

FOIL the right side:

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

$2 x^{2} + 10 x + 52 = x^{2} + 14 x + 49$ $2x^2+10x+52 = x^2 +14x +49$

Move the terms to one side of the equation:

$2 x^{2} + 10 x + 52 - x^{2} - 14 x - 49 = 0$ $2x^2+10x+52 - x^2-14x -49= 0$

combine like terms:

$x^{2} - 4 x + 3 = 0$ $x^2 -4x +3 =0$

factor:

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

solutions (roots):

$x = 1, x = 3$ $x = 1, x =3$

Answer 2 · 2018-05-22T00:34:46

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

Explanation:

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

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

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

$2 x^{2} + 10 x + 52 = x^{2} + 7 x + 7 x + 49$ $2x^2+10x+52=x^2+7x+7x+49$

$2 x^{2} + 10 x + 52 = x^{2} + 14 x + 49$ $2x^2+10x+52=x^2+14x+49$

$x^{2} + 10 x + 52 = 14 x + 49$ $x^2+10x+52=14x+49$

$x^{2} + 52 = 4 x + 49$ $x^2+52=4x+49$

$x^{2} + 3 = 4 x$ $x^2+3=4x$

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

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

$\frac{4 + \sqrt{16 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$ $(4+sqrt(16-4*1*3))/(2*1)$ or $\frac{4 - \sqrt{16 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$ $(4-sqrt(16-4*1*3))/(2*1)$

$\frac{6}{2} = 3$ $6/2=3$ or $\frac{2}{2} = 1 .$ $2/2=1.$

$x = 3 or x = 1$ $x=3 or x=1$

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

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