Solve (2x ^ 2 - 1) ^ 2 - 3x ^ 2 = 2(1/2 - 4x) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$(2x^{2}-1)^{2}-3x^{2}=2(\frac{1}{2}-4x)$$

Solve for x

$x = -\frac{9 ^ {\frac{2}{3}} {(\sqrt[3]{\sqrt{4155} + 72} + \sqrt[3]{72 - \sqrt{4155}})}}{18} \approx -1.708901045$
$x=0$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(2x^{2}-1\right)^{2}$.

$$4\left(x^{2}\right)^{2}-4x^{2}+1-3x^{2}=2\left(\frac{1}{2}-4x\right)$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$4x^{4}-4x^{2}+1-3x^{2}=2\left(\frac{1}{2}-4x\right)$$

Combine $-4x^{2}$ and $-3x^{2}$ to get $-7x^{2}$.

$$4x^{4}-7x^{2}+1=2\left(\frac{1}{2}-4x\right)$$

Use the distributive property to multiply $2$ by $\frac{1}{2}-4x$.

$$4x^{4}-7x^{2}+1=1-8x$$

Subtract $1$ from both sides.

$$4x^{4}-7x^{2}+1-1=-8x$$

Subtract $1$ from $1$ to get $0$.

$$4x^{4}-7x^{2}=-8x$$

Add $8x$ to both sides.

$$4x^{4}-7x^{2}+8x=0$$

Substitute $t$ for $x^{2}$.

$$4t^{2}-7t+8=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $4$ for $a$, $-7$ for $b$, and $8$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 4\times 8}}{2\times 4}$$

Do the calculations.

$$t=\frac{7±\sqrt{-79}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$t\in \emptyset$$

Since $t=x^{2}$, the original equation does not have any solutions.

$$x\in \emptyset$$