We have two set of solutions , ( $\frac{\sqrt{29} }{2}$ , $\frac{\sqrt{29} }{2}$ + 3 ) , ( - $\frac{\sqrt{29} }{2}$ + 3 , - $\frac{\sqrt{29} }{2}$ + 3).

<u>Step-by-step explanation:</u>

We have , the following equations:

$y = x+3$ and $y = x^{2} +2x -4$ . Let's substitute value of y in bottom equation:

⇒ $y = x^{2} +2x -4$

⇒ $x+3 = x^{2} +2x -4$

⇒ $x^{2} +x -7= 0$

Roots of quadratic equation are given by: a = 1, b= 1 , c = -7

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

⇒ $x = \frac{\sqrt{b^{2}-4ac} }{2a}$

⇒ $x = \frac{\sqrt{1^{2}-4(1)(-7)} }{2(1)}$

⇒ $x = \frac{\sqrt{29} }{2}$ which is actually x = ± $\frac{\sqrt{29} }{2}$ .

We know that $y = x+3$ ,

⇒ y = ± $\frac{\sqrt{29} }{2}$ + 3.

We have two set of solutions , ( $\frac{\sqrt{29} }{2}$ , $\frac{\sqrt{29} }{2}$ + 3 ) , ( - $\frac{\sqrt{29} }{2}$ + 3 , - $\frac{\sqrt{29} }{2}$ + 3).

5 0

3 years ago

Can someone please help me out?

djyliett [7]

Answer:

Correct answer is

$165 {c}^{6} {b}^{6}$