### Video Transcript

Which quadratic equation has roots π₯ equals two plus or minus π? (a) π₯ squared minus four π₯ plus five equals zero, (b) π₯ squared plus four π₯ plus five equals zero, (c) π₯ squared minus four π₯ plus three equals zero, (d) π₯ squared plus four π₯ plus three equals zero, or (e) π₯ squared minus five π₯ plus four equals zero.

In this question, weβve been given the roots of a quadratic equation and asked to find the equation itself. Now we could go through each of the suggested quadratic equations in turn and solve them using the quadratic formula. We could then identify which quadratic has roots which simplify to two plus π and two minus π. But there are five options, and thatβs quite a lengthy process, and there is a more efficient way.

If we know the roots of a quadratic equation, we can in fact work backwards to find the equation itself. We recall a general theorem. If the quadratic equation ππ₯ squared plus ππ₯ plus π equals zero has roots π₯ sub one and π₯ sub two, and π must be nonzero here, then the roots satisfy the following two properties. Firstly, π₯ sub one plus π₯ sub two is equal to negative π over π. And secondly, π₯ sub one multiplied by π₯ sub two is equal to π over π. In other words, the sum of the roots is equal to the negative of the coefficient of π₯ divided by the coefficient of π₯ squared. And the product of the two roots is equal to the constant term divided by the coefficient of π₯ squared.

If in fact the value of π is one, so the quadratic is of the form π₯ squared plus ππ₯ plus π equals zero, then these equations simplify to π₯ sub one plus π₯ sub two equals negative π and π₯ sub one multiplied by π₯ sub two equals π. We can demonstrate this by considering the quadratic equation π₯ squared plus seven π₯ plus 12 equals zero. To solve this equation, we would factor, and we would look for two numbers that add to the coefficient of π₯, thatβs positive seven, and multiply to the constant term. The two numbers that have a sum of seven and a product of 12 are three and four. So this quadratic factors as π₯ plus three multiplied by π₯ plus four is equal to zero.

The two roots of this quadratic equation are the values that make each set of parentheses in turn equal to zero. So they are negative three and negative four, which are the negatives of the two values inside the parentheses. The sum of these two roots, then, isnβt the value of π, but instead the negative of π. The sum is negative seven rather than positive seven. Their product though is equal to the constant term because weβve changed the sign of both from the values in the parentheses. So their product will still have the same magnitude and the same sign.

Now this illustrates but doesnβt prove the general result. It also gives us a clue as to how we can work backwards from knowing the roots of a quadratic equation to finding the equation itself. When deriving a quadratic equation from its roots, itβs easier to start with a simpler form, in which the coefficient of π₯ squared is equal to one. If either of the coefficients π or π turn out to be fractions, we can multiply the entire equation by the lowest common denominator of these two fractions to convert the quadratic equation to one with integer coefficients.

So starting with the quadratic equation π₯ squared plus ππ₯ plus π equals zero, we know that the sum of the roots will give the value negative π and the product will give the value π. Letβs define π₯ sub one to be the root two plus π and π₯ sub two to be the root two minus π. These are complex roots, but the result we stated above didnβt specify that the roots had to be real. And they donβt. It works regardless of whether the roots are real, complex, or purely imaginary.

Finding the sum of these two values, we have two plus π plus two minus π. The imaginary parts cancel out. And this simplifies to four. Remember that the sum of the roots gives the value of negative π. So we have the equation negative π equals four. We solve for π by multiplying or dividing both sides of this equation by negative one. And we find that π is equal to negative four.

To find the value of π, we need to take the product of these two roots. Using the FOIL method to distribute these parentheses gives four minus two π plus two π minus π squared. The imaginary terms in the center will cancel one another out. And at the same time, we recall that π squared is equal to negative one. So we have four minus negative one, thatβs four plus one, which is equal to five. The product of the roots, and therefore the value of π, is five.

All that remains is to substitute the values of π and π into the quadratic equation. So we have π₯ squared minus four π₯ plus five is equal to zero. Thatβs option (a) of the five we were given. We can check our answer by applying the quadratic formula to this equation only. The value of π is one, the value of π is negative four, and the value of π is five. So substituting these values into the quadratic formula, we have π₯ equals four plus or minus the square root of negative four squared minus four multiplied by one multiplied by five all over two multiplied by one.

The value under the square root evaluates to 16 minus 20, which is negative four. So these roots simplify to four plus or minus the square root of negative four all over two. The square root of negative four is the square root of four multiplied by the square root of negative one, which is two π. So we have four plus or minus two π all over two. And then dividing through by two, the roots simplify to two plus or minus π, which are the correct roots for the equation we were looking for.

We found then that the quadratic equation which has roots π₯ equals two plus or minus π is the equation π₯ squared minus four π₯ plus five is equal to zero.