MATH SOLVE

4 months ago

Q:
# Helllpppp plleeaassee (part 2)3. Consider the quadratic equation x^2 – 6x = –1.A. What is the value of the discriminant? Explain.B. How many solutions does the quadratic equation have and are those solutions rational, irrational, or nonreal? Explain.C. If the quadratic equation has real solutions, what are the solutions? Explain. Estimate irrational solutions to the nearest tenth.

Accepted Solution

A:

x²– 6x = –1

x²-6x+1=0, a=1, b=-6, c=1

A. Discriminant

D=b² - 4ac= 36-4*1*1=32, D>0, so this equation has 2 real solutions

to find the x we need to use formula

B.

[tex]x= \frac{-b+/- \sqrt{D} }{2a} \sqrt{D} = \sqrt{32} = \sqrt{(16*2)} =4 \sqrt{2} , \sqrt{D} =4 \sqrt{2} , irrational, so roots of the equation are going to be real irrational. [/tex]

C.x=(-b+/-√D)/2a, x1=(6+4√2)/2 = 3+2√2=5.8 , x2=(6-4√2)/2=3-2√2=0.2

x²-6x+1=0, a=1, b=-6, c=1

A. Discriminant

D=b² - 4ac= 36-4*1*1=32, D>0, so this equation has 2 real solutions

to find the x we need to use formula

B.

[tex]x= \frac{-b+/- \sqrt{D} }{2a} \sqrt{D} = \sqrt{32} = \sqrt{(16*2)} =4 \sqrt{2} , \sqrt{D} =4 \sqrt{2} , irrational, so roots of the equation are going to be real irrational. [/tex]

C.x=(-b+/-√D)/2a, x1=(6+4√2)/2 = 3+2√2=5.8 , x2=(6-4√2)/2=3-2√2=0.2