Question

Find a quadratic polynomial whose zeros are $5-3\sqrt{2}\text{ and 5+3}\sqrt{2}$.

Answer 1

Hint: Use the relation between the zeros and the coefficients of a polynomial. Also, try to draw some results seeing the nature of the roots.

Complete step-by-step answer:
 It is given that the roots of a quadratic equation are $5-3\sqrt{2}\text{ and 5+3}\sqrt{2}$.
We know, for standard quadratic equation $a{{x}^{2}}+bx+c=0$, the roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:
Sum of the roots of quadratic equation = $\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}$.
Product of the roots of quadratic equation = \[\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}\].
Now a quadratic equation with its roots given can be written as:
$k\left( {{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of roots} \right) \right)=0$
Where k is any constant and can have any value according to our need.
So, the quadratic equation with root as $5-3\sqrt{2}\text{ and 5+3}\sqrt{2}$ is:
 $k\left( {{x}^{2}}-\left( 5-3\sqrt{2}+5+3\sqrt{2} \right)x+\left( 5-3\sqrt{2} \right)\left( 5+3\sqrt{2} \right) \right)=0$
We know $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Therefore, our equation becomes:
$k\left( {{x}^{2}}-10x+{{5}^{2}}-{{\left( 3\sqrt{2} \right)}^{2}} \right)=0$
$\Rightarrow k\left( {{x}^{2}}-10x+25-18 \right)=0$
Now, as no condition given in the question, we let k to be 1. Therefore, our equation becomes: ${{x}^{2}}-10x+7=0$
Therefore, we can conclude that one of the quadratic equations with $5-3\sqrt{2}\text{ and 5+3}\sqrt{2}$ as its roots is ${{x}^{2}}-10x+7=0$.

Note: Remember that we can form infinite quadratic equations with both of its roots given, just we need to do is to keep on changing the value of k in the equation $k\left( {{x}^{2}}-\left( \text{sum of the roots} \right)x+\left( \text{product of roots} \right) \right)=0$. Also, remember that for a quadratic equation with integer coefficients as in the above case, irrational or complex roots always appear in conjugate pairs, i.e., if a+ib is one of the roots then the other root would be a-ib.