How do you solve #3x^4+12x^2-15=0#?

Question

Eva Abramson · Accepted Answer

Assuming we are restricted to Real (non-Complex) solutions:
#color(white)("XXX")x=+-1# Given #color(white)("XXX")3x^4+12x^2-15=0# Dividing both sides by #3# #color(white)("XXX")x^4+4x^2-5=0# Temporarily substituting #q# for #x^2# #color(white)("XXX")q^2+4q-5=0# Factoring the left side #color(white)("XXX")(q+5)(q-1) = 0# With potential solutions #color(white)("XXX")q=-5# and #q=1# If #q=-5# then #color(white)("XXX")x^2=-5# which is impossible for Real values of #x# If #q=1# then #color(white)("XXX")x^2=1# #rarrcolor(white)("X")x=+-1#

Abigail Allen · Answer

undefined To solve the equation 3x^4 + 12x^2 - 15 = 0:

1. Substitute $y = x^2$, so the equation becomes $3y^2 + 12y - 15 = 0$.
2. Solve this quadratic equation for $y$ using the quadratic formula: $y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
3. Substitute the solutions for $y$ back into $y = x^2$ to find the solutions for $x$.

Hazel Anglin · Answer

undefined To solve the equation $3x^4 + 12x^2 - 15 = 0$, you can use a substitution to simplify the equation. Let $y = x^2$, then the equation becomes a quadratic equation in terms of $y$. After solving for $y$, you can find the corresponding values of $x$ using the relation $y = x^2$.

Here's the process:

1. Substitute $y = x^2$ into the equation:
   $3(y^2) + 12y - 15 = 0$.

2. Solve the quadratic equation $3y^2 + 12y - 15 = 0$.

3. After finding the values of $y$, substitute them back into $y = x^2$ and solve for $x$.

Let's solve the quadratic equation $3y^2 + 12y - 15 = 0$:

1. Divide the equation by 3 to simplify it:
   $y^2 + 4y - 5 = 0$.

2. Factor the quadratic equation:
   $(y + 5)(y - 1) = 0$.

3. Set each factor equal to zero and solve for $y$:
   $y + 5 = 0 \implies y = -5$,
   $y - 1 = 0 \implies y = 1$.

Now, substitute back $y = x^2$ and solve for $x$:

For $y = -5$:
$x^2 = -5$ has no real solutions since the square of any real number cannot be negative.

For $y = 1$:
$x^2 = 1$ gives $x = \pm 1$.

Therefore, the solutions to the equation $3x^4 + 12x^2 - 15 = 0$ are $x = 1$ and $x = -1$.