Video: Pack 1 β€’ Paper 3 β€’ Question 22

Pack 1 β€’ Paper 3 β€’ Question 22

02:29

Video Transcript

Solve five π‘₯ squared minus three π‘₯ minus two is less than or equal to zero.

In order to be able to solve this inequality, we first need to solve five π‘₯ squared minus three π‘₯ minus two is equal to zero. To do this, we start by factorising fully. Since both five and two are prime numbers, we don’t have a huge number of options for the numbers inside our brackets.

First, we know that, at the front of each bracket, we must have a five π‘₯ and an π‘₯, since five π‘₯ multiplied by π‘₯ gives us five π‘₯ squared. We also know that the only factors of two are two and one.

To decide which order we choose, we can use a little trial and error. Five π‘₯ multiplied by one is five π‘₯. And two multiplied by π‘₯ is two π‘₯. Two π‘₯ minus five π‘₯ gives us the negative three we need and also ensures we get negative two rather than positive two as our constant.

Remember, for the product of this pair of brackets to be zero, either five π‘₯ plus two is equal to zero or π‘₯ minus one is equal to zero. For the first value of π‘₯, let’s start by subtracting two from both sides, then dividing by five. π‘₯ is equal to negative two-fifths. For the second value of π‘₯, we can just add one to both sides. π‘₯ is equal to one.

This last step is a little bit tricky. We sketch the curve of 𝑦 equals five π‘₯ squared minus three π‘₯ minus two. Remember, the π‘₯ values we worked out are the roots of the equation. They’re the points where the graph crosses the π‘₯-axis.

A quadratic curve with a positive coefficient of π‘₯ squared is a smiley face. The 𝑦-intercept is given by the constant negative two. Our inequality is five π‘₯ squared minus three π‘₯ minus two is less than or equal to zero. This is the part of the curve that lies on and below the π‘₯-axis. This occurs when π‘₯ is greater than or equal to negative two-fifths and when π‘₯ is less than or equal to one.

The values of π‘₯ that satisfy our inequality are, therefore, π‘₯ is greater than or equal to negative two-fifths and less than or equal to one.

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