Solve (x ^ 2 - 3) ^ 2 + x ^ 2 - 3 = 2 | 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

$$(x^{2}-3)^{2}+x^{2}-3=2$$

Solve for x

$x=-2$
$x=2$
$x=1$
$x=-1$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x^{2}-3\right)^{2}$.

$$\left(x^{2}\right)^{2}-6x^{2}+9+x^{2}-3=2$$

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

$$x^{4}-6x^{2}+9+x^{2}-3=2$$

Combine $-6x^{2}$ and $x^{2}$ to get $-5x^{2}$.

$$x^{4}-5x^{2}+9-3=2$$

Subtract $3$ from $9$ to get $6$.

$$x^{4}-5x^{2}+6=2$$

Subtract $2$ from both sides.

$$x^{4}-5x^{2}+6-2=0$$

Subtract $2$ from $6$ to get $4$.

$$x^{4}-5x^{2}+4=0$$

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

$$t^{2}-5t+4=0$$

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 $1$ for $a$, $-5$ for $b$, and $4$ for $c$ in the quadratic formula.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$t=\frac{5±3}{2}$$

Solve the equation $t=\frac{5±3}{2}$ when $±$ is plus and when $±$ is minus.

$$t=4$$ $$t=1$$

Since $x=t^{2}$, the solutions are obtained by evaluating $x=±\sqrt{t}$ for each $t$.

$$x=2$$ $$x=-2$$ $$x=1$$ $$x=-1$$