Question $\left(x^{2}-3\right)^{2}+3\left(x^{2}-3\right)-4=0$

Studdy Solution

STEP 1

1. The equation $\left(x^{2}-3\right)^{2}+3\left(x^{2}-3\right)-4=0$ is a quadratic equation in terms of the expression $x^{2}-3$ .
2. We can use substitution to simplify the equation and then solve for $x$ .
3. The solution should include all possible real values of $x$ .

STEP 2

1. Introduce a substitution to simplify the equation.
2. Solve the resulting quadratic equation.
3. Substitute back to find the values of $x$ .
4. Verify the solutions.

STEP 3

Introduce the substitution $u = x^2 - 3$ . Replace $x^2 - 3$ with $u$ in the equation.
$(u)^2 + 3(u) - 4 = 0$

STEP 4

Solve the quadratic equation $u^2 + 3u - 4 = 0$ using the quadratic formula, $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=1$ , $b=3$ , and $c=-4$ .
$u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}$ $u = \frac{-3 \pm \sqrt{9 + 16}}{2}$ $u = \frac{-3 \pm \sqrt{25}}{2}$ $u = \frac{-3 \pm 5}{2}$

STEP 5

Calculate the two possible values of $u$ .
$u_1 = \frac{-3 + 5}{2} = 1$ $u_2 = \frac{-3 - 5}{2} = -4$

STEP 6

Substitute back $u = x^2 - 3$ and solve for $x$ .
For $u_1 = 1$ : $x^2 - 3 = 1$ $x^2 = 4$ $x = \pm 2$
For $u_2 = -4$ : $x^2 - 3 = -4$ $x^2 = -1$ Since $x^2 = -1$ has no real solutions, ignore this case.

STEP 7

Verify the solutions $x = 2$ and $x = -2$ by substituting back into the original equation.
For $x = 2$ : $(2^2 - 3)^2 + 3(2^2 - 3) - 4 = (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0$
For $x = -2$ : $((-2)^2 - 3)^2 + 3((-2)^2 - 3) - 4 = (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0$
Both solutions satisfy the original equation.
The solutions to the equation $\left(x^{2}-3\right)^{2}+3\left(x^{2}-3\right)-4=0$ are $x = 2$ and $x = -2$ .

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