Question

Find the sum of the roots the following equation. The subject is quadratic equations with a maximum power of two (things like a(x^3)^2 + b(x^3) + c = 0 also count as a power of two) so we have to find a way to turn this into a quadratic equation. Please help.
$\left(2x\:^2-3x\right)^2+4\left(2x^2-3x\right)+3=0$

Answer 1

$\bf (2x^2-3x)^2+4(2x^2-3x)+3=0\qquad \boxed{a=2x^2-3x}\implies a^2+4a+3=0 \\\\\\ (a+3)(a+1)=0\implies a= \begin{cases} -3\\ -1 \end{cases}\qquad \stackrel{\textit{their sum}}{-3-1\implies -4}$

Answer 2

Answer:

3/2

Step-by-step explanation:

(2x² − 3x)² + 4 (2x² − 3x) + 3 = 0

We can simplify this using substitution.  Let's say that u = 2x² − 3x.

u² + 4u + 3 = 0

Factoring:

(u + 3)(u + 1) = 0

u = -3, u = -1

Therefore:

2x² − 3x = -3, 2x² − 3x = -1

2x² − 3x + 3 = 0, 2x² − 3x + 1 = 0

For a quadratic ax² + bx + c = 0, the sum of the roots is -b/a.  So in both cases, the sum is 3/2.