MATH SOLVE

2 months ago

Q:
# Which are the solutions of x2 = β5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 57 EndRoot Over 2 EndFraction StartFraction negative 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction negative 5 + StartRoot 7 EndRoot Over 2 EndFraction StartFraction 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 57 EndRoot Over 2 EndFraction StartFraction 5 minus StartRoot 7 EndRoot Over 2 EndFraction comma StartFraction 5 + StartRoot 7 EndRoot Over 2 EndFraction

Accepted Solution

A:

Answer:[tex]x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}[/tex]Step-by-step explanation:Given:The equation to solve is given as:[tex]x^2=-5x+8[/tex]Rearrange the given equation in standard form [tex]ax^2+bx +c =0[/tex], where, [tex]a,\ b,\ and\ c[/tex] are constants.Therefore, we add [tex]5x-8[/tex] on both sides to get,[tex]x^2+5x-8=0[/tex]Here, [tex]a=1,b=5,c=-8[/tex]The solution of the above equation is determined using the quadratic formula which is given as:[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]Plug in [tex]a=1,b=5,c=-8[/tex] and solve for [tex]x[/tex].[tex]x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}[/tex]Therefore, the solutions are:[tex]x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}[/tex]