We have two set of solutions , ( [tex]\frac{\sqrt{29} }{2}[/tex] , [tex]\frac{\sqrt{29} }{2}[/tex] + 3 ) , ( -[tex]\frac{\sqrt{29} }{2}[/tex] + 3 , -[tex]\frac{\sqrt{29} }{2}[/tex] + 3).
Step-by-step explanation:
We have , the following equations:
[tex]y = x+3[/tex] and [tex]y = x^{2} +2x -4[/tex] . Let's substitute value of y in bottom equation:
⇒ [tex]y = x^{2} +2x -4[/tex]
⇒ [tex]x+3 = x^{2} +2x -4[/tex]
⇒ [tex]x^{2} +x -7= 0[/tex]
Roots of quadratic equation are given by: a = 1, b= 1 , c = -7
[tex]x = \frac{\sqrt{b^{2}-4ac} }{2a}[/tex]
⇒ [tex]x = \frac{\sqrt{b^{2}-4ac} }{2a}[/tex]
⇒ [tex]x = \frac{\sqrt{1^{2}-4(1)(-7)} }{2(1)}[/tex]
⇒ [tex]x = \frac{\sqrt{29} }{2}[/tex] which is actually x = ± [tex]\frac{\sqrt{29} }{2}[/tex] .
We know that [tex]y = x+3[/tex],
⇒ y = ± [tex]\frac{\sqrt{29} }{2}[/tex] + 3.
We have two set of solutions , ( [tex]\frac{\sqrt{29} }{2}[/tex] , [tex]\frac{\sqrt{29} }{2}[/tex] + 3 ) , ( -[tex]\frac{\sqrt{29} }{2}[/tex] + 3 , -[tex]\frac{\sqrt{29} }{2}[/tex] + 3).
