Question 11(b) | BetaMath

Question 11(b)

Solve $x^2 - 8x - 9 \leq 0$

Solution

Rearrange the inequality to

\begin{aligned} x^2 - 8x - 9 &\leq 0 \\ (x^2 - 8x + 16) - 16 - 9 &\leq 0 \quad \text{We complete the square} \\ (x - 4)^2 - 25 &\leq 0 \\ (x - 4 - 5)(x - 4 + 5) &\leq 0 \quad \text{We factorise using the difference of perfect squares} \\ (x - 9)(x + 1) &\leq 0 \\ \end{aligned}

The answer is the interval between the two values of $x$ which satisfy the equality in this relation, so our answer is

x \in [-1, 9]

Explanation

What is the question asking us to do

In this question, we see that this is an equality that involves a quadratic expression.

If we drew the two sides of this equation, the left hand side would be a parabola, and the right hand side is exactly $0$ so this would be the x-axis.

Since we are looking for the left hand side to be the lesser in this inquality, this question is asking us to find the values of $x$ which define the section of the parabola which is below the x-axis. It is important that we remember to include the points which are on the x-axis, since this is a weak-inequality (rather than a strict inequality).

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