Answer:
70/9
Step-by-step explanation:
We have the quadratic:
[tex]3x^2+4x-9[/tex]
So, let’s find the roots of the quadratic. We will set the expression equal to 0:
[tex]3x^2+4x-9=0[/tex]
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:
[tex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case:
[tex]a=3, b=4,\text{ and } c=-9[/tex]
Therefore, by substitution:
[tex]\displaystyle x=\frac{-(4)\pm\sqrt{(4)^2-4(3)(-9)}}{2(3)}[/tex]
Evaluate:
[tex]\displaystyle x=\frac{-4\pm\sqrt{124}}{6}[/tex]
Simplify the square root:
[tex]\sqrt{124}=\sqrt{4\cdot31}=2\sqrt{31}[/tex]
Hence:
[tex]\displaystyle x=\frac{-4\pm2\sqrt{31}}{6}[/tex]
Reduce:
[tex]\displaystyle x=\frac{-2\pm\sqrt{31}}{3}[/tex]
So, our roots are:
[tex]\displaystyle x_1=\frac{-2+\sqrt{31}}{3}, x_2=\frac{-2-\sqrt{31}}{3}[/tex]
We want to find the sum of the squares of our two roots. So, let’s square each term:
[tex]\displaystyle (x_1)^2=\Big(\frac{-2+\sqrt{31}}{3}\Big)^2[/tex]
Square. For the numerator, we can use the perfect square trinomial patten where:
[tex](a+b)^2=(a^2+2ab+b^2)[/tex]
Therefore:
[tex]\displaystyle (x_1)^2=\Big(\frac{(-2)^2+2(-2)(\sqrt{31})+(\sqrt{31})^2}{9}\Big)[/tex]
Simplify:
[tex]\displaystyle (x_1)^2=\frac{35-4\sqrt{31}}{9}[/tex]
Similarly, for the second root, we will have:
[tex]\displaystyle (x_2)^2=\Big(\frac{-2-\sqrt{31}}{3}\Big)^2[/tex]
So:
[tex]\displaystyle (x_2)^2=\Big(\frac{(-2)^2+2(-2)(-\sqrt{31})+(-\sqrt{31})^2}{9}\Big)[/tex]
Simplify:
[tex]\displaystyle (x_2)^2=\frac{35+4\sqrt{31}}{9}[/tex]
Therefore, our sum will be:
[tex]\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}[/tex]
Therefore, our final answer is 70/9.
