+0

# Algebra

0
4
3
+1420

Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.

Jun 27, 2024

#1
+1230
+1

First, let's combine all like terms for x and move all terms to one side. We have that

$$x^{2}+12x-4=0$$

Using the quadratic equation, we have that

$$x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}$$

$$x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6$$

Now, we can use a handy trick to solve this problem. We have that

$$(2\sqrt{10}-6)^2+(-2\sqrt{10}-6)^2$$

Expanding and distributing in our values, we have

$$-24\sqrt{10}+76+4\left(\sqrt{10}\right)^{2}+12\sqrt{10}-6\left(-2\sqrt{10}-6\right)$$

All the radicals basically cancel out. In the end, we are left with 152.

Thanks! :)

Jun 27, 2024
#2
+14
+1

Simplifying we get $$x^2 + 12x - 4 = 0$$. Using the quadratic formula we find the roots to be $$-6 + 2\sqrt{10}$$ and $$-6 - 2\sqrt{10}$$. Taking their squares, we get $$36 + 4\sqrt{10} + 40$$ and $$36 - 4\sqrt{10} + 40$$. Adding them together, we get $$\boxed{152}$$

Jun 27, 2024
#3
+129725
+1

Let the roots be a,b

Product of the roots  = ab =  -4   →  2ab = -8

Sum of the roots =  a + b = -12     square both sides

a^2 + b^2 + 2ab = 144

a^2 + b^2 - 8 =144

a^2 + b^2 =  152

Jun 27, 2024