Video Transcript
Find the quadratic equation with
real coefficients which has five plus π as one of its roots.
Weβre 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.
Weβre going to need to distribute
these parentheses. Letβs use the grid method here
since thereβs 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 weβll 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
thatβs five π β and the inner two terms β thatβs 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 weβre 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 ππ, the equation of that quadratic
is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero. π is the real part of the
solution. Here thatβs five. And π is the imaginary part. In our solution, thatβs 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.