Question

Find the quadratic equation whose roots are $9-\sqrt{5},9+\sqrt{5}$
A. $x^2-18x-56=0$
B. $x^2+18x+56=0$
C. $4x^2-72x-56=0$
D. $x^2-18x+76=0$

Answer 1

Hint: An equation which has two roots or of degree 2 is called a quadratic equation. If the roots of a quadratic equation are ‘a’ and ‘b’, then that quadratic equation in variable x will be in the form $x^2-\left(a+b\right)x+ab=0$ . So add and multiply the given roots and substitute them in this general form accordingly to get the required equation.
Formula used:
 $\left(a+b\right)\left(a-b\right)=a^2-b^2$

Complete step-by-step answer:
We are given to find the quadratic equation whose roots are $9-\sqrt{5},9+\sqrt{5}$ .
Let $9-\sqrt{5}$ be ‘a’ and $9+\sqrt{5}$ be ‘b’.
When the roots of a quadratic equation are ‘a’ and ‘b’, then the equation will be $x^2-\left(a+b\right)x+ab=0$
Where $a+b$ is the sum of the roots and $ab$ is the product of the roots of the quadratic equation
 $a+b=\left(9-\sqrt{5}\right)+\left(9+\sqrt{5}\right)=9+9-\sqrt{5}+\sqrt{5}$
Cancelling the similar terms with different signs, we get $a+b=9+9=18$
 $ab=\left(9-\sqrt{5}\right)\left(9+\sqrt{5}\right)$
The RHS of the above equation is in the form $\left(x-y\right)\left(x+y\right)$ which is equal to $x^2-y^2$
Therefore, $ab=9^2-{\left(\sqrt{5}\right)}^2$
 $\Rightarrow ab=81-5=76$
On substituting the values of obtained $a+b$ and $ab$ in $x^2-\left(a+b\right)x+ab=0$ , we get
 $\Rightarrow x^2-\left(18\right)x+76=0$
Therefore the quadratic equation whose roots are $9-\sqrt{5},9+\sqrt{5}$ is $x^2-18x+76=0$
So, the correct answer is “Option D”.

Note: No. of roots of an equation is determined by its highest degree. If the highest degree is 3, then the equation will have 3 roots, if 4 then the equation will have 4 roots and so on. An equation can have both real roots and imaginary roots.