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]
[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]
