solving..(x^2-5x-1)^2-18(x^2-5x-1)+65=0?

Question

(x^2-5x-1)^2-18(x^2-5x-1)+65=0

how do you solve this im lost.

Answer 1

Substituting x^2-5x-1 by 'a'
The given expression
=a^2-18a+65=0
a^2-13a-5a+65=0
a(a-13)-5(a-13)=0
(a-13)(a-5)=0
a=13 or 5
if a=13,then
x^2-5x-1=13 [putting back the value of a]
x^2-5x-14=0
(x+2)(x-7)=0
x= -2 or 7
If a=5,then
x^2-5x-1=5
x^2-5x-6=0
(x+1)(x-6)=0
x= -1 or 6
Hence the values of x are-2,or7 or-1 or6

Answer 2

The first term is the messy one. One easy way to handle it is to do it like an old fashioned arithmetic multiplication problem, as I show below:
x^2 -5x -1
x^2 -5x -1
________
x^4 -5x^3 -x^2
......-5x^3+25x^2 +5x
...............-x^2 +5x + 1
__________________
x^4 -10x^3+23x^2 +10x + 1

I hope this is a useful technique for you. Having got the first term worked out, the second term is easy, so we can put all this together to get

x^4-10x^3+23x^2+10x+1-18x^2+90x=18+65=0
x^4-10x^3+5x^2+100x+83=0
What a mess ! Let's go back to the beginning-there must be an easier way. And indeed there is.

(x^2-5x-1)^2 -18(x^2-5x-1)+65 = 0
(x^2-5x-1)[(x^2-5x-1) - 18] +65 =0
Let (x^2-5x-1)=Z
Z(Z-18)+65 = 0
Z^2 - 18Z + 65 =0
(Z-15)(Z-3) =0, so Z=15 or Z=3 Aha!
Now replace Z by (x^2-5x-1), and we get
x^2 -5x-1 =15, or 3
If x^2 -5x -1=15, x^2 -5x -16 =0
If x^2 -5x -1 =3, x^2 -5x -4 =0

Neither of these quadratics is easily factorable, so you have to use the Quadratic formula' The first
equation, x^2-5x-16 will solve to give
x= [5 +or- sq.rt.(25+64)]/2, = [5+or-sqrt(69)]/2
The second equation gives x=[5+or-sqrt(25+16)]/2
which gives [5+or-sqrt(41)]/2

That's the best I can do for you. I hope it helps.
Good luck

Answer 3

5x+1/3 - 2x+1/4=7/12 First, I combined the x's together to get 3x 3x + 1/3 + 1/4 = 7/12 Next, I converted the fractions on the left side to 12th's 3x + 4/12 + 3/12 = 7/12 Next, I added the fractions together on the left side 3x + 7/12 = 7/12 Then, I subtracted 7/12 from both sides 3x = 0 Then I divided both sides by 3 x = 0 I try making solving equations like that as simple and easy as I can. Normally that can done by first simpliflying wherever you can simplify. That's why I worked with the x's first. Then another way I could simplify more was if I got all of the fractions with a common denominator. Then it just kind of falls into place after that.

Answer 4

(x^2-5x-1)^2-18(x^2-5x-1)+65=0
Let z = x^2-5x-1
Then z^2-18z+65 = 0
(z-13)(z-5) = 0
So z = 5 or 13
Thus x^2 -5x-1 = 5
x^2-5x-6= 0
(x-6)(x+1) = 0 so x = 6 or -1
x^2-5x-1 = 13
x^2-5x -14 =0
(x-7)(x+2) = 0 so x = 7 or - 2
Thus there are 4 values of x: -1,-2,6, and 7