Question

question
solve the equation for all real solutions in simplest form.
$b^2 - 17b + 16 = -2b^2$

Explanation:

Step1: Rearrange to standard quadratic form

Add $2b^2$ to both sides:
$b^2 + 2b^2 - 17b + 16 = 0$
Simplify:
$3b^2 - 17b + 16 = 0$

Step2: Apply quadratic formula

For $ax^2+bx+c=0$, roots are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=3$, $b=-17$, $c=16$:
$\text{Discriminant } D = (-17)^2 - 4(3)(16) = 289 - 192 = 97$
Roots: $\frac{17\pm\sqrt{97}}{2\times3}$

Answer:

$b=\frac{17+\sqrt{97}}{6}$ and $b=\frac{17-\sqrt{97}}{6}$