Question

In need of some help! Thank you in advance. 100 points <3

1: Explain why the equation (x-4)^2-17=8 has two solutions. Then solve the equation to find the solutions. Show your work.

2: This is Josh’s solution for the equation x^2+4x+3=0:

x^2+4x+3=0
x^2+4x=-3
x^2+4x+4=-3+4
(x+2)^2=1
x+2=1
x=-1

Is Josh’s solution correct? Explain.

3:Use the quadratic formula to solve x^2-6x+7. Show your work. Then describe the solution. No work, no credit.

Answer 1

1

the equation (x-4)^2-17=8 has two solutions because it is in a quadratic form.

(x-4)^2-17=8

x²-8x+16-17-8=0

x²-8x-9=0

Doing middle term factorization

x²-9x+x-9=0

x(x-9)+1(x-9)=0

(x-9)(x+1)=0

either

x=9

x=9or

x=9orx=-1

2.

x^2+4x+3=0:

Doing middle term factorization

x²+3x+x+3=0

x(x+3)+1(x+3)=0

(x+3)(x+1)=0

either

x=-3

or

x=-1

Josh’s solution is incorrect because he missed x=-3.

3.

x^2-6x+7=0

By using quadratic equation formula:

Comparing above equation with ax²+bx+c=0

we get

a=1

b=-6

c=7

Now

we have

x=
`\frac{ - b ± \sqrt{ {b}^(2) - 4ac} }{2a}`

x=
`\frac{ 6± \sqrt{ {-6}^(2) - 4*1*7} }{2}`

x=
`( 6±2√( 2))/(2)`

x=
`{ 3±√( 2)}`

Taking positive

x=
`{ 3+√( 2)}`

taking negative

x=
`{ 3-√( 2)}`