Question

(iii) $x^4 - 2x^2 - 63 \leq 0$

Answer 1

[-3, 3]

Answer 2

The given inequality is $x^4 - 2x^2 - 63 \leq 0$. We can rewrite this inequality by treating it as a quadratic in terms of $x^2$. Let $y = x^2$. Since $x$ is a real number, $y = x^2 \geq 0$. Substituting $y$ into the inequality, we get: $y^2 - 2y - 63 \leq 0$

Now, we solve this quadratic inequality for $y$. 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 expression $(y - 9)(y + 7)$ represents a parabola opening upwards. It is less than or equal to zero between its roots. So, the inequality $(y - 9)(y + 7) \leq 0$ is satisfied when: $-7 \leq y \leq 9$

Now, substitute back $y = x^2$: $-7 \leq x^2 \leq 9$

This compound inequality can be split into two separate inequalities:

1. $x^2 \geq -7$
2. $x^2 \leq 9$

Let's analyze the first inequality, $x^2 \geq -7$. For any real number $x$, its square $x^2$ is always non-negative, i.e., $x^2 \geq 0$. Since $0 \geq -7$, the inequality $x^2 \geq -7$ is true for all real numbers $x$.

Now, let's analyze the second inequality, $x^2 \leq 9$. This inequality is satisfied when the absolute value of $x$ is less than or equal to the square root of 9: $|x| \leq \sqrt{9}$ which means $|x| \leq 3$, so $-3 \leq x \leq 3$.

We need the values of $x$ that satisfy both $x^2 \geq -7$ and $x^2 \leq 9$. Since $x^2 \geq -7$ is true for all real $x$, the solution set is determined solely by the inequality $x^2 \leq 9$. The solution to $x^2 \leq 9$ is $-3 \leq x \leq 3$.

Alternatively, after factoring the original expression directly: $x^4 - 2x^2 - 63 = (x^2)^2 - 2(x^2) - 63$. This can be factored as a quadratic in $x^2$: $(x^2 - 9)(x^2 + 7) \leq 0$

For any real number $x$, $x^2 \geq 0$, so $x^2 + 7 \geq 7 > 0$. The factor $(x^2 + 7)$ is always positive. For the product $(x^2 - 9)(x^2 + 7)$ to be less than or equal to zero, the other factor $(x^2 - 9)$ must be less than or equal to zero: $x^2 - 9 \leq 0$, which means $x^2 \leq 9$. Taking the square root of both sides, we get $|x| \leq 3$, which means $-3 \leq x \leq 3$.

The solution is the set of all real numbers $x$ such that $-3 \leq x \leq 3$. In interval notation, this is $[-3, 3]$.