(x^2+5x)^2+2(x^2+5x)-24=0

Question

find all real solutions 

 

Answer 1

let's call x² + 5x = a, so then the equation becomes:

 

a² + 2a - 24 = 0 This factors into:

 

(a + 6)(a - 4) = 0, so a = -6 or 4  Now replace a with x² + 5x 

 

So x² + 5x = -6  or  x² + 5x = 4 

 

Add 6 to the first one and subtract 4 from the second one:

 

x² + 5x + 6 = 0 or  x² + 5x - 4 = 0

 

The first one factors: 

(x + 3)(x + 2) = 0,  so x = -3, -2

 

The second one won't factor, so we can complete the square to give   x + 5/2 = ±√41/2, so x = (-5 ± √41)/2 

 

 

You would expect 4 answers because it is a degree 4 equation.

Answer 2

Notice that the term x² + 5x shows up twice.  Let's do a substitution for it and see what happens.

 

Let y = x² + 5x.  Now the equation is:  y² + 2y - 24 = 0.

 

That can be solved for y, by factoring:  (y + 6) (y - 4) = 0

 

So y = -6 or y = 4.

 

Now go back to the original substitution.  The solutions for y let us say:

 

x² + 5x = -6 or x² + 5x = 4.

 

That gives us two more quadratic equations that we can solve:

 

EQUATION 1

x² + 5x = -6

x² + 5x + 6 = 0

(x + 3) (x + 2) = 0

x = -3, or x = -2

 

EQUATION 2

x² + 5x = 4

x² + 5x - 4 = 0

 

This one can't be factored, so we have to use the quadratic formula:

 

x = [-5 ± √(5² - 4(1)(-4))] / 2

x = [-5 ± √(25 + 16)] / 2

x = [-5 ± √41] / 2

 

So there are 4 solutions:

 

x = -3, x = -2, x = (-5 + √41) / 2, x = (-5 - √41) / 2

 

 

Answer 3

Let x^2+5x=t.  Then t^2+2t-24=0.  Use quadratic equation to solve for t.  Then remember that t=x^2+5x.   Solve this again using quadratic equation.