# Question Video: Solving Quadratic Equations by Quadratic Formula Mathematics • First Year of Secondary School

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Use the quadratic formula to find all the values of π for which the roots of the equation 2π₯Β² + ππ₯ β 7 = 0 differ by exactly 10. Give your answers correct to two decimal places if necessary.

07:42

### Video Transcript

Use the quadratic formula to find all the values of π for which the roots of the equation two π₯ squared plus ππ₯ minus seven equals zero differ by exactly 10. Give your answers correct to two decimal places if necessary.

In this question, weβre given a quadratic equation with an unknown coefficient of π₯ as π. We need to find the values of π for which the roots of this equation differ by exactly 10. Donβt forget that the roots of an equation are simply the values of π₯ here, for which two π₯ squared plus ππ₯ minus seven is equal to zero. Weβre told to use the quadratic formula, so letβs recall what the quadratic formula is. It tells us that if we have an equation in the general form ππ₯ squared plus ππ₯ plus π is equal to zero, where π, π, and π are constants and π is not equal to zero, then π₯ is equal to negative π plus or minus the square root of π squared minus four ππ all over two π.

The equation that we were given is in this general form. However, weβll need to be a little careful as the unknown coefficient of π₯ which was given as π is different to the coefficient of π₯ squared in the general form. But weβll do what we should always do and clearly write out the values of π, π, and π ready to plug into the formula. π is the coefficient of π₯ squared, so thatβs two. π is the coefficient of π₯, so thatβs π. And π is negative seven. So we write out the quadratic formula and fill in the values. So weβll have π₯ is equal to negative π plus or minus the square root of π squared minus four times two times negative seven over two times two.

When we simplify this, we can work out that four times two times negative seven gives us negative 56. And when we subtract negative 56 from π squared, then we end up with π squared plus 56 within the square root. On the denominator, two times two gives us four. At this point, we can return to the information in the question, which tells us that the roots of this equation differ by exactly 10. Remember that the quadratic formula will give us two solutions and they come from this plus or minus symbol. In this question, weβll have π₯ is equal to negative π plus the square root of π squared plus 56 over 4 and π₯ is equal to negative π minus the square root of π squared plus 56 over 4.

When the roots of this equation differ by exactly 10, that means that the larger solution for π₯ subtract the smaller solution for π₯ will be equal to 10. Letβs clear some space so we can do some working. So here we have the two solutions for π₯, which weβre subtracting as weβre finding the difference. And weβve set this difference equal to 10. Now we can solve this equation to find the value of π. The first thing we can do is to realize that on the left-hand side we have two fractions and theyβre both written as fractions over four. This means that the answer will be a fraction over four and everything on the numerator of the second fraction must be subtracted from that of the first fraction.

When we distribute this negative across the parentheses, weβll have plus π and plus the square root of π squared plus 56. We can now notice that we have negative π plus π, which is equal to zero. And we also have two lots of the square root of π squared plus 56. We can then write this equation in the form two times the square root of π squared plus 56 over 4 is equal to 10. We can also take out a common factor of two from the numerator and denominator. The next step in our calculation would be to multiply both sides of the equation by two, which gives us that the square root of π squared plus 56 is equal to 20.

Next, we can square both sides of this equation and see that π squared plus 56 is equal to 400. Subtracting 56 from both sides, we can see that π squared is equal to 344. Finally, we can take the square root of both sides of this equation. But as we need to consider both the positive and negative values of the square root, we can use the plus or minus sign. As we need to give our answer correct to two decimal places, that means we can use our calculator to find the decimal equivalent of the square root of 344. So π is equal to 18.5472 and so on. And when we round to two decimal places, we check the third decimal digit to see if itβs five or more. As it is, then we can say that π is approximately equal to 18.55 to two decimal places.

But of course we know that negative 18.5472 and so on squared would also give us 344. So that means that negative 18.55 is also a solution for π to two decimal places. We can, therefore, give the answer that the values of π for which the roots of the equation differ by exactly 10 are π is equal to 18.55 and negative 18.55 to two decimal places. Of course, itβs always worthwhile checking the values. We started with the quadratic equation two π₯ squared plus ππ₯ minus seven equals zero. And we had to work out the values of π. But when weβre checking the values of π that we calculated, it would be a good idea to use the square root of 344 rather than the decimal approximations that we found as the answer.

So when the roots of this equation differ by exactly 10, weβll have two π₯ squared plus the square root of 344π₯ minus seven equals zero. We can then use the quadratic formula with the values of π equals two, π equals the square root of 344, and π equals negative seven. When we begin to simplify what we have, the square root of 344 squared is, of course, 344. Then we need to multiply four times two times negative seven, which is negative 56. And when we subtract that, we can remember that weβve previously worked this out. 344 plus 56 would give us 400.

Taking the square root of 400 gives us 20. So weβre left with π₯ is equal to negative the square root of 344 plus or minus 20 over four. The first solution when π₯ is equal to negative root 344 plus 20 over 4 gives us that π₯ is equal to 0.363 and so on. The second solution, when we have π₯ is equal to negative root 344 minus 20 over four, is π₯ is equal to negative 9.636 and so on.

So what exactly does all that checking achieve? Well, if we look at the two solutions for π₯, what is the difference between these? Itβs 10, which demonstrates that if we have the equation two π₯ squared plus ππ₯ minus seven equals zero and the roots of the equation differ by exactly 10, then the values of π must be 18.55 and negative 18.55 to two decimal places.

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