Question

solve the following quadratic equation for all values of x in simplest form. 2(x + 7)^2+41 = 43

Explanation:

Step1: Isolate the squared term

Subtract 41 from both sides of the equation $2(x + 7)^2+41 = 43$.
$2(x + 7)^2=43 - 41$
$2(x + 7)^2=2$

Step2: Solve for $(x + 7)^2$

Divide both sides of the equation by 2.
$(x + 7)^2=\frac{2}{2}$
$(x + 7)^2 = 1$

Step3: Take the square - root of both sides

$x+7=\pm\sqrt{1}$
$x + 7=\pm1$

Step4: Solve for x

Case 1: When $x + 7 = 1$
Subtract 7 from both sides: $x=1 - 7=-6$
Case 2: When $x + 7=-1$
Subtract 7 from both sides: $x=-1 - 7=-8$

Answer:

$x=-6$ or $x=-8$