Question

Please help me! Due soon!
The quadratic $3x^2+4x-9$ has two real roots. What is the sum of the squares of these roots? Express your answer as a common fraction in lowest terms.

Answer 1

Answer:

70/9

Step-by-step explanation:

We have the quadratic:

$3x^2+4x-9$

So, let’s find the roots of the quadratic. We will set the expression equal to 0:

$3x^2+4x-9=0$

Testing for factors, we can see that our quadratic isn’t factorable.

So, we can use the Quadratic Formula. The quadratic formula is given by:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case:

$a=3, b=4,\text{ and } c=-9$

Therefore, by substitution:

$\displaystyle x=\frac{-(4)\pm\sqrt{(4)^2-4(3)(-9)}}{2(3)}$

Evaluate:

$\displaystyle x=\frac{-4\pm\sqrt{124}}{6}$

Simplify the square root:

$\sqrt{124}=\sqrt{4\cdot31}=2\sqrt{31}$

Hence:

$\displaystyle x=\frac{-4\pm2\sqrt{31}}{6}$

Reduce:

$\displaystyle x=\frac{-2\pm\sqrt{31}}{3}$

So, our roots are:

$\displaystyle x_1=\frac{-2+\sqrt{31}}{3}, x_2=\frac{-2-\sqrt{31}}{3}$

We want to find the sum of the squares of our two roots. So, let’s square each term:

$\displaystyle (x_1)^2=\Big(\frac{-2+\sqrt{31}}{3}\Big)^2$

Square. For the numerator, we can use the perfect square trinomial patten where:

$(a+b)^2=(a^2+2ab+b^2)$

Therefore:

$\displaystyle (x_1)^2=\Big(\frac{(-2)^2+2(-2)(\sqrt{31})+(\sqrt{31})^2}{9}\Big)$

Simplify:

$\displaystyle (x_1)^2=\frac{35-4\sqrt{31}}{9}$

Similarly, for the second root, we will have:

$\displaystyle (x_2)^2=\Big(\frac{-2-\sqrt{31}}{3}\Big)^2$

So:

$\displaystyle (x_2)^2=\Big(\frac{(-2)^2+2(-2)(-\sqrt{31})+(-\sqrt{31})^2}{9}\Big)$

Simplify:

$\displaystyle (x_2)^2=\frac{35+4\sqrt{31}}{9}$

Therefore, our sum will be:

$\displaystyle (x_1)^2+(x_2)^2\\\\ \begin{aligned} &=\frac{35-4\sqrt{31}}{9}+\frac{35+4\sqrt{31}}{9}\\&=\frac{35-4\sqrt{31}+35+4\sqrt{31}}{9}\\&=\frac{70}{9}\end{aligned}$

Therefore, our final answer is 70/9.