y2 = x2 + 5x – 27; x + 5 = y2.How many real solutions are there to the system of equations above?

Question

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Answer 1

x²+5x-27=x+5=y², so x²+5x-x-27-5=0, x²+4x-32=(x+8)(x-4)=0, and x=-8 or 4.

y²=x+5=-8+5=-3 or y²=4+5=9, so y=±3.  Two real solutions appear to be (x,y)=(4,3) and (4,-3), but we need to test them first:

x=4: y²=x²+5x-27⇒9=16+20-27=9 (true) for y=-3 or 3. There is no real solution for y when x=-8, so there are only two real solutions for (x,y).

 

Answer 2

x^2 + 5x -27 = x+5

x^2 + 4x - 32 = 0

B^2 - 4ac:  4^2 -4(1)(-32) = 144

144>0 if x is greater then 0 there are 2 real roots