Question

solve:

15x^4-28x^2=-5

Answer 1

\[\Huge x^2 (15 x^2-28) = -5 \]

Answer 2

(3x^2-5)(5x^2-5)=0
x^2=5/3 -----> x = sqrt(5/3) 0r x= -sqrt(4/3)
x^2=1 ----> x=1 or x=-1

Answer 3

15*x^4 -28*x^2 +5 = 0 =>
=> 15*x^4 -28*x^2 +5 = 0
=> 15*(x +1.291)*(x +0.447)*(x -0.447)*(x -1.291) = 0

Answer 4

\[15x^4 -28x^2 = -5\] What you should do is use substitution, then the quadratic formula, then substitute again.

Here we go:

Allow \[y=x^2\]

Now, substitute "y" in the equation by replacing instances of x^2 with "y". We will replace x^4 with y^2 because \[y^2 = (x^2)^2 = x^4\]

The equation now is
\[15y^2 - 28y =-5\] and rearrange this to get \[15y^2 -28y+5=0\] and we can now use the quadratic formula to solve for y because the equation is now of the form \[ax^2 + bx +c =0\]
Thus, \[y=\frac{-b \pm \sqrt{b^2 - 4*a*c}}{2a}\] and we have that \[a = 15\]\[b=28\]\[c=5\]
Once we have solved for "y" here, we can now solve for x by using the fact that \[y=x^2\] so \[x=\pm\sqrt{y}\]