Question

Solve the following inequality:

$x^{4}-2x^{2}-63 \leq 0$

Answer 1

[-3, 3]

Answer 2

Let $y = x^2$. Since $x$ is a real number, $x^2 \geq 0$, so $y \geq 0$. Substituting $y=x^2$ into the inequality, we get:

$y^2 - 2y - 63 \leq 0$

This is a quadratic inequality in terms of $y$. To solve it, we first find the roots of the corresponding quadratic equation $y^2 - 2y - 63 = 0$. We can factor the quadratic expression:

$(y - 9)(y + 7) = 0$

The roots are $y = 9$ and $y = -7$. The quadratic $y^2 - 2y - 63$ represents a parabola opening upwards. It is less than or equal to zero between its roots. So, the inequality $y^2 - 2y - 63 \leq 0$ is satisfied when:

$-7 \leq y \leq 9$

Now, we must consider the condition $y \geq 0$ that came from the substitution $y = x^2$. We need the values of $y$ that satisfy both $-7 \leq y \leq 9$ and $y \geq 0$. The intersection of the interval $[-7, 9]$ and the interval $[0, \infty)$ is $[0, 9]$. So, we have:

$0 \leq y \leq 9$

Now, substitute back $y = x^2$:

$0 \leq x^2 \leq 9$

This inequality can be split into two separate inequalities:

1. $x^2 \geq 0$
2. $x^2 \leq 9$

For the first inequality, $x^2 \geq 0$, this is true for all real numbers $x$. For the second inequality, $x^2 \leq 9$, this is equivalent to taking the square root of both sides and considering the absolute value:

$\sqrt{x^2} \leq \sqrt{9}$

$|x| \leq 3$

The inequality $|x| \leq 3$ means that $-3 \leq x \leq 3$. We need the values of $x$ that satisfy both $x^2 \geq 0$ and $x^2 \leq 9$. Since $x^2 \geq 0$ is true for all real $x$, the solution to $0 \leq x^2 \leq 9$ is simply the solution to $x^2 \leq 9$. Thus, the solution to the inequality $x^2 \leq 9$ is $-3 \leq x \leq 3$. In interval notation, the solution set is $[-3, 3]$.