Video: AQA GCSE Mathematics Higher Tier Pack 2 β€’ Paper 1 β€’ Question 18

Circle the two roots of the equation (π‘₯ βˆ’ 3) (4 βˆ’ 3π‘₯) = 0. [A] βˆ’3 [B] 3 [C] 4/3 [D] βˆ’4/3.


Video Transcript

Circle the two roots of the equation π‘₯ minus three multiplied by four minus three π‘₯ equal zero. The four possibilities are negative three, three, four-thirds, and negative four-thirds.

The equation that we’ve been given is a quadratic equation because if we were to expand the brackets, then the highest power of π‘₯ that would appear in the equation would be two, which would arise when we multiply π‘₯ by negative three π‘₯, giving negative three π‘₯ squared.

The roots of a quadratic equation are the solution to that equation. They’re the values of π‘₯ that make the equation true. Now we have two brackets which multiply together to give zero. And in order to get zero when we multiply two things together, it must be true that at least one of those individual things, in this case brackets, is equal to zero. So we can set each of the brackets equal to zero in turn and then solve the resulting linear equations to find the values of π‘₯ that satisfy these equations.

The first equation, π‘₯ minus three equals zero, can be solved by adding three to each side. This will cancel the negative three on the left. And on the right, we’ll now have zero plus three, which is three. So this equation solves to give π‘₯ equals three. This is therefore our first root of the given equation. But notice that the question says we need to circle the two roots.

The second linear equation could be solved in two ways. But as we have a negative number of π‘₯s on the left of the equation, the easiest thing to do will be to add three π‘₯ to each side so that we now have a positive number of π‘₯s on the right. Adding three π‘₯ to the left will cancel with a negative three π‘₯ and just leave four. And adding three π‘₯ on the right gives three π‘₯.

So now we have four equals three π‘₯. To find the value of π‘₯, we need to divide both sides of the equation by three, giving four-thirds equals π‘₯. So the second root of this equation is four-thirds.

Another meaning of the roots of a quadratic equation is that they are the π‘₯-intercepts of the graph of 𝑦 equals that quadratic function. If we were to sketch the graph of 𝑦 equals π‘₯ minus three multiplied by four minus three π‘₯, it would be a negative or 𝑛-shaped parabola, because the coefficient of π‘₯ squared would be negative.

Remember, we said that it will be equal to negative three. The values at which this curve would cross the π‘₯-axis are four-thirds and three, which are the two roots of the quadratic equation.

Notice that the other two roots we were given as possible options, so that’s negative three and negative four-thirds, were actually the same size as the actual roots. But the signs were negative rather than positive. This is because a common mistake, certainly in the case of the first root at least, would be to just write down the values inside the brackets, so negative three rather than three. But as we’ve seen, this is incorrect. And we actually need to set each bracket equal to zero and then solve in order to find the roots.

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