Video: AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 17

AQA GCSE Mathematics Higher Tier Pack 1 β€’ Paper 3 β€’ Question 17

06:13

Video Transcript

Solve four π‘₯ squared minus two π‘₯ minus seven equal zero using the quadratic formula. Give your answer to three significant figures.

We’ve been given a quadratic equation, that’s one with the highest power of π‘₯ is two, and specifically asked to solve this equation by using the quadratic formula. Let’s recall its definition. The quadratic formula tells us how to solve the general quadratic equation π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equal zero. The roots or the solutions to this quadratic equation are π‘₯ equals minus 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž.

Remember, π‘Ž, 𝑏, and 𝑐 represent the coefficients in the quadratic equation. π‘Ž is the coefficient of π‘₯ squared; 𝑏 is the coefficient of π‘₯; and 𝑐 is the constant term. In our quadratic equation, the coefficient of π‘₯ squared is four, so this is the value of π‘Ž. the coefficient of π‘₯ is negative two, so this is the value of 𝑏. The constant term, so that’s the one without any π‘₯ s or π‘₯ squareds, is negative seven, so this is the value of 𝑐. You must make sure that you look at the signs of each coefficient carefully.

Now that we’ve identified the values of π‘Ž, 𝑏, and 𝑐 for this quadratic equation, we can substitute them into the quadratic formula, taking great care with each of the signs. First, we have π‘₯ equal minus 𝑏. So this is minus negative two. Then plus or minus the square root of 𝑏 squared, so this is plus or minus square root of negative two squared. We need to be extra careful with this part. We need the brackets around all of negative two. Negative two all squared is negative two multiplied by negative two, which is equal to four as a negative multiplied by negative gives a positive. But if we leave out the brackets and just write negative two squared, then this actually means negative one multiplied by two squared. And if we remember BIDMAS, or the order of operations, then powers or indices come before multiplication. So we’d work out two squared first, which is four, and then multiply it by negative one, giving negative four.

We get a different answer if we leave out the brackets compared to if we leave them in. So this is particularly important if you’re typing this into your calculator. You must make sure that you include brackets around negative two. Continuing under the square root, we then have minus four π‘Žπ‘, so this means minus four multiplied by π‘Ž multiplied by 𝑐. So we have minus four multiplied by four multiplied by negative seven. And again I’ve included all of this within brackets. The square root sign extends over the whole of this expression. In the denominator of the fraction, we have two multiplied by π‘Ž which is equal to two multiplied by four.

Now you can just type this all into your calculator in one go. And if you’re careful with negative signs and brackets as we’ve already discussed, then you will get the correct answer. But it’s sensible to break the calculation down into stages, not least because there might be marks available for showing subsequent stages of this calculation. So simplifying this then, first of all, we have minus negative two, which is just equal to two as the two minus signs together form a positive. Under the square root, we first have negative two all squared, which remember is equal to four not negative four as already discussed. Four multiplied by four multiplied by negative seven is equal to negative 112. So under the square root, we have four minus negative 112.

Again be very careful with signs here. A common mistake would just be to write down four minus 112. And in the denominator, we have two multiplied by four, which is equal to eight. The square root can be simplified further because four minus negative 112 is just equal to four plus 112. Four plus 112 is 116. So our values of π‘₯ simplify to two plus or minus the square root of 116 all over eight. Now there is some further simplification that could go on here. For example, we could simplify two over eight to one quarter or we could simplify the surd of square root of 116. But we’ve been asked to give answers to three significant figures, so we need to evaluate the two roots as decimals.

For the first root, we work out two plus the square root of 116 over eight, and it gives a decimal 1.59629. For the second root, we work out two minus the square root of 116 over eight, and this time it gives negative 1.09629. The question asked us to give our answer to three significant figures. So in each case we need to look at the fourth significant figure in order to decide whether we’re rounding up or down. For the first root, the fourth significant figure is a six, so we’re rounding up. But the nine in the next column will have to round up to ten, so we round the five in the next column up to six, and this root rounds to 1.60.

For the second root, the fourth significant figure is also a six, and again there’s a nine in the next column. So we round the nine up to ten and this root becomes negative 1.10. So the solutions to the quadratic equation four π‘₯ squared minus two π‘₯ minus seven equal zero, each to three significant figures, are 1.60 or negative 1.10. We do need to include the zero in the second decimal place each time, although it doesn’t actually affect the size of the number, because we’ve been specifically asked to give our answer to three significant figures. And each of these zero are the third significant figure.

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