# Question Video: Reconstructing a Quadratic Equation from a Nonreal Root Mathematics • 10th Grade

Find the quadratic equation with real coefficients which has 5 + π as one of its roots.

03:24

### Video Transcript

Find the quadratic equation with real coefficients which has five plus π as one of its roots.

Weβre told that five plus π is a root of the quadratic equation. And remember, we know that the nonreal roots of a quadratic equation which has real coefficients occur in complex conjugate pairs. To find the complex conjugate, we change the sign of the imaginary part. And we can therefore see that the roots of our equation are five plus π and five minus π. And this means that our quadratic equation is of the form π₯ minus five plus π multiplied by π₯ minus five minus π is equal to zero. And this comes from the fact that when we solve a quadratic equation by factoring, we equate each expression inside the parentheses to zero.

So in this case, we would have π₯ minus five plus π equals zero and π₯ minus five minus π equals zero. We would solve this first equation by adding five plus π to both sides. And we see that π₯ equals five plus π. And we solve the second equation by adding five minus π to both sides. And we get that second root π₯ equals five minus π.

Weβre going to need to distribute these parentheses. Letβs use the grid method here since thereβs a number of bits and pieces that could trip us up. π₯ multiplied by π₯ is π₯ squared. π₯ multiplied by negative five plus π is negative π₯ five plus π. Similarly, we get negative π₯ multiplied by five minus π. And negative five plus π multiplied by negative five minus π gives us a positive five minus π multiplied by five plus π. And weβll distribute these brackets using the FOIL method.

Multiplying the first term in the first bracket by the first term in the second bracket gives us 25. We multiply the outer two terms β thatβs five π β and the inner two terms β thatβs negative five π. And five π minus five π is zero. So these cancel each other out. And then we multiply the last terms. Negative π multiplied by π is negative π squared. And since π squared is equal to negative one, we see that these brackets distribute to be 25 minus negative one, which is equal to 26. So our quadratic equation is currently of the form π₯ squared minus π₯ multiplied by five plus π minus π₯ multiplied by five minus π plus 26.

We collect like terms. And we get π₯ squared minus five plus π plus five minus π multiplied by π₯ plus 26. π minus π is zero. And weβre left with π₯ squared minus 10π₯ plus 26 equals zero.

Now actually there is a formula that we can use that will save us some time. If we have a quadratic equation with real roots and a compact solution π plus ππ, the equation of that quadratic is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero. π is the real part of the solution. Here thatβs five. And π is the imaginary part. In our solution, thatβs one.

We can substitute what we know about our complex number into the formula. And we get π₯ squared minus two times five times π₯ plus five squared plus one squared. Two times five is 10, and five squared plus one squared is 26. And we see once again we have the same quadratic equation. And we should be able to see now why this method can be a bit of a time saver.