SOLUTION: Can someone please help me with these? Please? Give exact and approximate solutions to three decimal places. x^2-7x+9=0 x^2+8x+16=49 (x-8)^2=81

Question 469000: Can someone please help me with these? Please?
Give exact and approximate solutions to three decimal places.
x^2-7x+9=0
x^2+8x+16=49
(x-8)^2=81
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
x2-7x+9=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=13 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5.30277563773199, 1.69722436226801. Here's your graph:

-----------------------------
x2+8x+16=49
x2+8x-33=0
(x+11)(x-3)
x=-11 or 3
-----------------------------
(x-8)2=81
x2-16x+64=81
x2-16x-17=0
(x+1)(x-17)=0
x=-1 or 17..
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