Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 3 • Question 22

(a) Write 𝑥² − 12𝑥 − 14 in the form (𝑥 − 𝑎)² − 𝑏. (b) A quadratic equation whose turning point is (5, 3) has the equation 𝑦 = 𝑥² + 𝑘𝑥 + 𝑙. Find the values of 𝑘 and 𝑙.

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Video Transcript

Part a) Write 𝑥 squared minus 12𝑥 minus 14 in the form 𝑥 minus 𝑎 all squared minus 𝑏. Part b) A quadratic equation whose turning point is five, three has the equation 𝑦 equals 𝑥 squared plus 𝑘𝑥 plus 𝑙. Find the values of 𝑘 and 𝑙.

The method of writing any quadratic expression in the form 𝑥 minus 𝑎 all squared minus 𝑏 is known as completing the square. Our first step when completing the square is to divide the coefficient of 𝑥 by two. In this case, we need to divide negative 12 by two. This is equal to negative six. Our second step is to subtract the square of this number. In this case, we need to subtract negative six squared. Negative six multiplied by negative six is equal to 36 as multiplying a negative number by a negative number gives us a positive answer. this leaves us with 𝑥 minus six all squared minus 36.

We now need to drop the negative 14 into this line. Negative 36 minus 14 is equal to negative 50. This means that 𝑥 squared minus 12𝑥 minus 14 written in the form 𝑥 minus 𝑎 all squared minus 𝑏 is equal to 𝑥 minus six all squared minus 50. For completeness, we could write that 𝑎 is equal to six and 𝑏 is equal to 50. We could also check this answer by expanding the bracket subtracting 50. And our end result would be 𝑥 squared minus 12𝑥 minus 14, the original expression.

The second part of the question told us that a quadratic equation had a turning point five, three and it had an equation of 𝑦 equals 𝑥 squared plus 𝑘𝑥 plus 𝑙. The turning point can either be a maximum or a minimum point. In this case, it will be a minimum. As the coefficient of 𝑥 squared in the equation is positive, the graph will be U-shaped.

Any quadratic equation written in the form 𝑥 minus 𝑎 all squared plus 𝑏 has a turning point at the coordinates 𝑎, 𝑏. As our equation has a turning point five, three, we can substitute 𝑎 equals five and 𝑏 equals three into this equation. This gives us 𝑦 is equal to 𝑥 minus five all squared plus three. 𝑥 minus five all squared is the same as 𝑥 minus five multiplied by 𝑥 minus five.

We can expand these double brackets using the FOIL method. Multiplying the first terms 𝑥 multiplied by 𝑥 gives us 𝑥 squared. Multiplying the outside terms gives us negative five 𝑥. Multiplying the inside terms also gives us negative five 𝑥. Finally, multiplying the last terms gives us positive 25 as negative five multiplied by negative five is equal to 25. Simplifying by collecting the like terms gives us 𝑥 squared minus 10𝑥 plus 25.

We can therefore say that 𝑦 is equal to 𝑥 squared minus 10𝑥 plus 25 plus three. 25 plus three is equal to 28. Therefore, 𝑦 is equal to 𝑥 squared minus 10𝑥 plus 28. Our value for 𝑘 is the coefficient of 𝑥 and our value for 𝑙 is the constant. This means that 𝑘 is equal to negative 10 and 𝑙 is equal to 28.

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