Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract

Question

14 days ago

8

Which are the solutions of x2 = –5x + 8? StartFraction negative 5 minus StartRoot 57 EndRoot Over 2 EndFraction comma StartFract

ion 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

Mathematics

2 answers:

tester [3.9K] · Answer 1 · 2026-02-18T21:04:59+03:00

Answer:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Detailed solution:

Given:

The problem to solve is:

$x^2=-5x+8$

Convert the equation into the standard quadratic form $ax^2+bx +c =0$ , where $a,\ b,\ and\ c$ represent constants.

So, by adding $5x-8$ to both sides, we get:

$x^2+5x-8=0$

Note that $a=1,b=5,c=-8$ .

The roots of this quadratic are found by applying the quadratic formula given as:

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

Substitute $a=1,b=5,c=-8$ into the formula and calculate for $x$ .

$x=\frac{-5\pm \sqrt{5^2-4(1)(-8)}}{2(1)}\\x=\frac{-5\pm \sqrt{25+32}}{2}\\x=\frac{-5\pm \sqrt{57}}{2}\\\\\\\therefore x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Hence, the roots are:

$x=\frac{-5-\sqrt{57} }{2}\ or\ x=\frac{-5+\sqrt{57} }{2}$

Zina [3.9K] · Answer 2 · 2026-02-18T21:08:35+03:00

Zina [3.9K]14 days ago

3 0

Answer:

The correct choice is B. Hope that clarifies it!

Step-by-step explanation: