Question

Solve for a -8=a(2+5)^2+41

Answer 1

Greetings!

Solve for a:
$-8=a(2+5)^2+41$

Distribute the Parenthesis:
$-8=(2a+5a)^2+41$

Apply the Exponent:
$-8=(2a+5a)(2a+5a)+41$

Use Binomial Multiplication:
$-8=2a(2a+5a)+5a(2a+5a)+41$

$-8=(4a^2+10a^2)+(10a^2+25a^2)+41$

Combine Like Terms:
$-8=4a^2+10a^2+10a^2+25a^2+41$

$-8=49a^2+41$

Add -41 to both sides:
$(-8)+(-41)=(49a^2+41)+(-41)$

$-49=49a^2$

Divide both sides by 49:
$\frac{-49}{49}= \frac{49a^2}{49}$

$-1=a^2$

Sqaure Root both sides:
$\sqrt{-1}= \sqrt{a^2}$

$-1=a$

$a=-1$

The Answer Is:
$\boxed{a=-1}$

I hope this helped!
-Benjamin