Video: Pack 4 β€’ Paper 3 β€’ Question 13

Pack 4 β€’ Paper 3 β€’ Question 13

03:12

Video Transcript

Write π‘₯ squared plus six π‘₯ minus two in the form π‘₯ plus π‘Ž all squared plus 𝑏 where π‘Ž and 𝑏 are integers.

Writing a quadratic expression in this form is called completing the square. As with factorizing, there are lots of methods to get to the correct answer. In this case, we will look at two.

Our first method will be a step-by-step process that will work every time the coefficient of π‘₯ squared is equal to one. This means there is no number in front of the π‘₯ squared term. Our first step using this method is to divide the coefficient of π‘₯ by two. In this case, six divided by two is equal to three. Our next step is to subtract the square of the number we have just written inside the bracket. In this case, we’re going to subtract three squared. Finally, we dropped the last term into the next line, in this case negative two.

Simplifying this gives us π‘₯ plus three all squared minus nine minus two. And finally, grouping the last two terms gives us π‘₯ plus three all squared minus 11. Our expression has now been written in the correct form, where π‘Ž is equal to three and 𝑏 is equal to negative 11.

An alternative method that you might have been taught is that any quadratic expression π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 can be rewritten as π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž, where 𝑝 is equal to 𝑏 divided by two π‘Ž and π‘ž is equal to 𝑐 minus 𝑏 squared divided by four π‘Ž. In our expression π‘₯ squared plus six π‘₯ minus two, our values of π‘Ž, 𝑏, and 𝑐 are one, six, and negative two, respectively. 𝑝 can, therefore, be calculated by dividing six by two multiplied by one. This is equal to three.

π‘ž is equal to negative two minus six squared divided by four multiplied by one. Six squared is equal to 36 and four multiplied by one is equal to four. 36 divided by four is equal to nine, which leaves us with π‘ž is equal to negative two minus nine. This is equal to negative 11. As π‘Ž is equal to one, substituting in our values of 𝑝 and π‘ž gives us the same answer.

π‘₯ squared plus six π‘₯ minus two can be rewritten as π‘₯ plus three all squared minus 11.

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