Lesson Video: Forming a Quadratic Equation Using Another Quadratic Equation | Nagwa Lesson Video: Forming a Quadratic Equation Using Another Quadratic Equation | Nagwa

# Lesson Video: Forming a Quadratic Equation Using Another Quadratic Equation Mathematics • First Year of Secondary School

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In this video, we will learn how to write a quadratic equation given the roots of another quadratic equation.

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### Video Transcript

In this lesson, weβll learn how to write a quadratic equation given the roots of another quadratic equation. Letβs begin by recalling the relationship between the roots of a quadratic and the coefficients of its terms.

For a quadratic equation of the form ππ₯ squared plus ππ₯ plus π equals zero, by the quadratic formula, its solutions, and as its roots are π one and π two, are π one equals negative π plus the square root of π squared minus four ππ over two π. And π two is equal to negative π minus the square root of π squared minus four ππ over two π. Weβre now going to find the sum and product of these general roots.

The sum is π one plus π two. The expression for the sum is shown. But of course, we can split the first fraction up into negative π over two π plus the square root of π squared minus four ππ over two π. And then, we can split the second fraction into negative π over two π minus the square root of π squared minus four ππ over two π. And now, we see that the second term and the fourth term sum to make zero; essentially, they cancel. And so, we have negative π over two π plus negative π over two π, which is negative two π over two π. Finally, we can cancel by dividing both the numerator and denominator of our fraction by two, leaving us with negative π over π. So, the sum of our roots is negative π over π. In other words, itβs the negative of the coefficient of π₯ over the coefficient of π₯ squared.

Weβre now going to repeat this process to find the product. Product means multiply, so weβre doing π one times π two. Once again, we split the first fraction up, and then we split the second fraction up. And weβre now going to distribute by multiplying the first terms, the outer terms, the inner terms, and the last terms. Multiplying the first terms gives us π squared over four π squared. Then, multiplying the outer and inner terms gives us the positive and negative versions of π times the square root of π squared minus four ππ all over four π squared. Now, of course, the sum of these is zero. So, these terms are actually going to cancel.

Then, when we multiply the last terms, we get negative π squared minus four ππ all over four π squared. Weβre now going to split this second fraction up, remembering, of course, that since weβre going to be subtracting negative four ππ over four π squared, thatβs the same as adding it. And so we should see now that π squared over four π squared minus π squared over four π squared is zero. Similarly, we can divide through by an π and by the four in this final fraction, and that leaves us simply with π over π. And we can therefore say that the product of the two roots of a quadratic equation is the constant term divided by the coefficient of π₯ squared.

Now whilst this might not seem like a hugely useful result, if we go back to our original equation ππ₯ squared plus ππ₯ plus π equals zero and divide through by π β weβre allowed to do this, of course, cause π is not going to be equal to zero β we see that our quadratic equation becomes π₯ squared plus π over ππ₯ plus π over π equals zero. And now, we should see that these link to our earlier results. Negative π over π is the same as the negative coefficient of π₯, and π over π is the constant term. And so, weβre now able to say that a quadratic equation whose leading coefficient is one can be written as π₯ squared minus the sum of roots times π₯ plus the product of the roots equals zero.

So, letβs have a look at an example of how we might apply these results.

Given that πΏ plus three and π plus three are the roots of the equation π₯ squared plus eight π₯ plus 12 equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ and π.

Letβs begin by recalling the relationship between a quadratic equation whose leading coefficient is one and its roots. We can write the equation in the form π₯ squared minus that sum of the roots times π₯ plus the product of the roots equals zero. And so, if we consider our equation π₯ squared plus eight π₯ plus 12 equals zero, the sum of the roots must be negative eight. Now, the reason itβs negative eight and not positive eight is because in the general form, the coefficient of π₯ is the negative sum of the roots, and in our example, the coefficient is positive. We can also say that the product, which is the constant term, must be equal to 12.

And often weβll look to replace the roots with the roots given in the question. So here the roots are πΏ plus three and π plus three. But in fact, weβre able to calculate two numbers which have a sum of negative eight and a product of 12. Two numbers which satisfy this result are negative six and negative two, and this is because a negative times a negative is a positive. So, negative six times negative two is positive 12. But also, negative six plus negative two is negative eight. And so, defining our roots to be π one and π two, we see that theyβre equal to negative six and negative two.

But of course, we were told that the roots of this equation are πΏ plus three and π plus three. So we can form two equations, one in πΏ and one in π. The first equation is πΏ plus three equals negative six, and the second is π plus three equals negative two. Weβll solve both of these equations for their respective variable by subtracting three from both sides. Negative six minus three is negative nine. So πΏ must be equal to negative nine. Similarly, π is equal to negative five. And so this is really useful because we now know the roots to our new equation. We can form this equation by finding the sum of these roots and the product. The sum is negative nine plus negative five, which is negative 14. And the product is negative nine times negative five, which is 45.

And so now that we know that the sum of our roots is negative 14 and the product is 45, we can substitute these back into the general form. That gives us π₯ squared minus negative 14 times π₯ plus 45 equals zero, and this simplifies to π₯ squared plus 14π₯ plus 45 equals zero.

Weβre now going to look at a similar example. But this time, weβre not going to be able to easily work out the roots of our original equation, and so weβre going to use algebra to find the new equation.

Given that πΏ and π are the roots of the equation π₯ squared minus two π₯ plus five equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ squared and π squared.

Letβs begin by recalling the relationship between a quadratic equation whose leading coefficient is one and its roots. We can represent it as π₯ squared minus the sum of the roots times π₯ plus the product of the roots equals zero. And so, this is essentially saying that if we have a quadratic equation equal to zero and the leading coefficient is one, in other words, the coefficient of π₯ squared is one, the negative coefficient of π₯ will tell us the sum of the roots and the constant term will tell us its product.

So we take the equation π₯ squared minus two π₯ plus five equals zero. The coefficient of π₯ is negative two. And so the sum must be the negative of negative two, so the sum of the roots must be positive two. Then the constant term is five. So the product of the roots must be five. So can we find two numbers that have a sum of two and a product of five? Well, no, not easily. Weβre not going to get nice integer solutions. And so instead, weβre going to form equations using πΏ and π. Since the sum of our roots is two and πΏ and π are the roots, we can say πΏ plus π must be equal to two. And then we can say that πΏ times π is equal to five.

The roots of our new equation are πΏ squared and π squared. And so, since their sum will be πΏ squared plus π squared, we need to manipulate our equations to find an expression for πΏ squared plus π squared and another expression for their product, πΏ squared π squared. Letβs label our equations as one and two. Weβre going to take the entirety of equation one and, weβre going to square it. In other words, we square both sides. So on the right-hand side, we get two squared, which is, of course, equal to four. Then, on the left-hand side, we get πΏ plus π squared, which we can consider to be πΏ plus π times πΏ plus π.

And if we distribute these parentheses, we get πΏ squared plus two πΏπ plus π squared equals four. And then, if we subtract two πΏπ from both sides, we get the expression for the sum of the roots πΏ squared and π squared. Itβs four minus two πΏπ. But of course, we have an expression for πΏπ; equation two tells us that πΏπ is equal to five. And so, πΏ squared plus π squared becomes four minus two times five, which is four minus 10 or simply negative six. So, we found the sum of the roots of our new equation and, therefore, the negative coefficient of π₯.

Weβre now going to repeat this process for equation two; weβre going to square both sides. That is, πΏπ squared equals five squared. But, of course, five squared is 25. And we can distribute to the power of two across both terms. And we get πΏ squared π squared equals 25. We, therefore, find that the sum of our new roots is negative six and the product is 25. Letβs substitute these back into the general form. When we do, we get π₯ squared minus negative six π₯ plus 25 equals zero. And so, the quadratic equation whose roots are πΏ squared and π squared is π₯ squared plus six π₯ plus 25 equals zero.

In our next example, weβll look at using the relationship between the coefficients of a quadratic and its roots to help us find the value of an expression.

If πΏ and π are the roots of the equation π₯ squared plus 20π₯ plus 15 equals zero, what is the value of one over π plus one over πΏ?

We begin by reminding ourselves what the relationship between the coefficient of a quadratic equation is and its roots. For a quadratic equation whose leading coefficient is one, in other words, the coefficient of π₯ squared is one, the negative coefficient of π₯ tells us the sum of the roots and the constant term tells us the product. And this is really useful because the coefficient of π₯ here is 20 and the constant is 15. And so, the sum of our roots must be negative 20. Remember, we said that itβs the negative coefficient of π₯. Then, the product, which is the constant term, must be 15.

But of course, we were told that πΏ and π are the roots of our equation. So we can, in fact, say that πΏ plus π must be negative 20 and πΏ times π must be 15. So, how does this help? Weβre looking to find the value of one over π plus one over πΏ, and we canβt easily find two numbers that have a sum of negative 20 and a product of 15. So weβre going to need to manipulate our expressions. Letβs think of one over π plus one over πΏ.

We know that to add two fractions, we need to create a common denominator. Now, the easiest way to do this when working with algebraic fractions is to multiply both parts of each fraction by the denominator of the other. So, weβre going to multiply the numerator and denominator of our first fraction by πΏ and of our second fraction by π. That gives us πΏ over πΏπ plus π over πΏπ. And now since the denominators are the same, we simply add the numerators, and that gives us πΏ plus π over πΏπ.

And this is really useful because we know that the numerator πΏ plus π is equal to negative 20, and then the denominator πΏπ is 15. And this means, in turn, that one over π plus one over πΏ is equal to negative 20 over 15, which simplifies to negative four-thirds. And so, if πΏ and π are the roots of our equation, then one over π plus one over πΏ must be equal to negative four-thirds.

Weβre now going to extend this idea into forming quadratic equations.

Given that πΏ and π are the roots of the equation three π₯ squared plus 16π₯ minus one equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ over two and π over two.

Letβs begin by recalling the relationship between the quadratic equation and its roots. For a quadratic equation whose leading coefficient is one, we can say that the coefficient of π₯ is the negative sum of the roots of the equation, whereas the constant is the product of its roots. Now, comparing this to our equation, we see we do have a bit of a problem. The leading coefficient, the coefficient of π₯ squared, is three. And so, weβre going to divide every single term by three. Three π₯ squared divided by three is π₯ squared. 16π₯ divided by three is 16π₯ over three or 16 over three π₯. And then, our constant term becomes negative one-third.

And so, comparing this to the general form, we know that the sum of the roots of our equation is the negative coefficient of π₯. So, itβs negative 16 over three. Then, the product is negative one-third. But weβre also told that πΏ and π are the roots of our equation, so we can replace these with πΏ plus π as the sum and πΏ times π as the product. Weβre looking to find the quadratic equation whose roots are πΏ over two and π over two. And so, what weβre going to do is find an expression for the sum of these roots; thatβs πΏ over two plus π over two. Similarly, weβre going to find the product of these roots; thatβs πΏ over two times π over two which is πΏπ over four.

So, how can we link the equations we have? Well, letβs call this first equation one. We have πΏ plus π as being equal to negative 16 over three. If we divide the entire expression by two, that is πΏ plus π over two, we actually know thatβs equal to πΏ over two plus π over two. And so, this means we can find the value of πΏ over two plus π over two by dividing the value for πΏ plus π by two. Thatβs negative 16 over three divided by two, which is negative eight over three.

And we can repeat this process with our second equation. This time, of course, we want πΏπ divided by four. So thatβs going to be negative one-third divided by four, which is negative one twelfth. And now that we have the sum of our roots and the product, we can substitute these back into our earlier equation. When we do, we find that the quadratic equation whose roots are πΏ over two and π over two is π₯ squared minus negative eight over three π₯ plus negative one twelfth equals zero.

Letβs simplify a little by dealing with our signs. In other words, negative negative eight-thirds is just eight-thirds. And adding negative one twelfth is the same as subtracting one twelfth. Our final step is to create integer coefficients. And to do so, weβre going to multiply through by 12. π₯ squared times 12 is 12π₯ squared. Then, if we multiply eight-thirds by 12, we cancel a three. And we end up working out eight times four, which is 32. Negative one twelfth times 12 is negative one. And, of course, zero times 12 is zero. And so, the quadratic equation is 12π₯ squared plus 32π₯ minus one equals zero.

Letβs now recap the key points from this lesson. In this video, we learned that for an equation of the form ππ₯ squared plus ππ₯ plus π equals zero, the sum of the roots is negative π over π. Itβs the negative coefficient of π₯ divided by the coefficient of π₯ squared. And the product is π over π. Itβs the constant term divided by the coefficient of π₯ squared. We can use this information to express the relationship between the roots of a quadratic whose leading coefficient is one and the coefficient of its terms. When the equation is written in this form, the negative coefficient of π₯ tells us the sum of the roots, whereas the constant term tells us the product.

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