MATH SOLVE

2 months ago

Q:
# What is the solution set of x^{2} +5x-5=0?

Accepted Solution

A:

We can find the solution using the quadratic formula:

[tex]x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a} [/tex]

b =coefficient of x term = 5

a = coefficient of squared term = 1

c = constant term = -5

Using the values in the above formula we get:

[tex]x= \frac{-5+- \sqrt{25-4(1)(-5)} }{2(1)} \\ \\ x= \frac{-5+- \sqrt{45} }{2} \\ \\ x= \frac{-5+-3 \sqrt{5} }{2} \\ \\ x= \frac{-5+3 \sqrt{5} }{2}, x= \frac{-5-3 \sqrt{5} }{2} [/tex]

[tex]x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a} [/tex]

b =coefficient of x term = 5

a = coefficient of squared term = 1

c = constant term = -5

Using the values in the above formula we get:

[tex]x= \frac{-5+- \sqrt{25-4(1)(-5)} }{2(1)} \\ \\ x= \frac{-5+- \sqrt{45} }{2} \\ \\ x= \frac{-5+-3 \sqrt{5} }{2} \\ \\ x= \frac{-5+3 \sqrt{5} }{2}, x= \frac{-5-3 \sqrt{5} }{2} [/tex]