College

Solve: [tex]42x^4 - 45x^2 + 12 = 0[/tex].

Enter all solutions separated by a comma.

Do not include "x =" in your answer.

Answer :

We start with the equation

[tex]$$42x^4 - 45x^2 + 12 = 0.$$[/tex]

Notice that the equation is quadratic in terms of [tex]$x^2$[/tex]. To simplify the process, we substitute

[tex]$$u = x^2,$$[/tex]

so that the equation becomes

[tex]$$42u^2 - 45u + 12 = 0.$$[/tex]

### Step 1. Solve the Quadratic Equation in [tex]$u$[/tex]
We use the quadratic formula

[tex]$$u = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A},$$[/tex]

with

[tex]$$A = 42,\quad B = -45,\quad C = 12.$$[/tex]

First, calculate the discriminant:

[tex]$$\Delta = B^2 - 4AC = (-45)^2 - 4(42)(12) = 2025 - 2016 = 9.$$[/tex]

Then, the solutions for [tex]$u$[/tex] are

[tex]$$u = \frac{45 \pm \sqrt{9}}{2 \cdot 42} = \frac{45 \pm 3}{84}.$$[/tex]

This gives us two solutions:

1. When using the plus sign,

[tex]$$u_1 = \frac{45 + 3}{84} = \frac{48}{84} \approx 0.5714285714285714.$$[/tex]

2. When using the minus sign,

[tex]$$u_2 = \frac{45 - 3}{84} = \frac{42}{84} = 0.5.$$[/tex]

### Step 2. Return to [tex]$x$[/tex] by Taking Square Roots
Recall that [tex]$u = x^2$[/tex]. Therefore, for each value of [tex]$u$[/tex], we have

[tex]$$x = \pm \sqrt{u}.$$[/tex]

For [tex]$u_1 \approx 0.5714285714285714$[/tex]:

[tex]$$x = \pm \sqrt{0.5714285714285714} \approx \pm 0.7559289460184544.$$[/tex]

For [tex]$u_2 = 0.5$[/tex]:

[tex]$$x = \pm \sqrt{0.5} \approx \pm 0.7071067811865476.$$[/tex]

### Final Answer
The four solutions to

[tex]$$42 x^4-45 x^2+12=0$$[/tex]

are

[tex]$$0.7559289460184544, -0.7559289460184544, 0.7071067811865476, -0.7071067811865476.$$[/tex]