Question

(i) Find the equation of a quadratic polynomial whose zeroes are $2\sqrt{3}$ and $-2\sqrt{3}$?
(ii) Find the value of k if -4 is the zero of the quadratic polynomial ${{x}^{2}}-x-\left( 2k+2 \right)$?

Answer 1

Hint: We start solving the problem (i) by recalling the sum and product of the zeros of the quadratic polynomial. We then assume the equation of the quadratic polynomial and then find the sum and product of the given roots. We then use these values to find the equation of the quadratic polynomial. We then solve problem (ii) by recalling the definition of zero of a polynomial. We then substitute the given zero in the polynomial and equate it to zero. We then make necessary calculations to find the value of k.

Complete step-by-step answer:
(i) According to the problem, we need to find the equation of a quadratic polynomial whose roots are $2\sqrt{3}$ and $-2\sqrt{3}$.
We know that the general equation of the quadratic polynomial is $a{{x}^{2}}+bx+c$. We know that sum and product of the zeroes of the quadratic polynomial $a{{x}^{2}}+bx+c$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$.
Now, let us assume the equation of the quadratic polynomial to be ${{x}^{2}}+bx+c$ ---(1).
So, we get $-b=2\sqrt{3}+\left( -2\sqrt{3} \right)$.
$\Rightarrow -b=2\sqrt{3}-2\sqrt{3}$.
$\Rightarrow -b=0$ ---(2).
We also have \[c=\left( 2\sqrt{3} \right)\times \left( -2\sqrt{3} \right)\].
$\Rightarrow c=-{{\left( 2\sqrt{3} \right)}^{2}}$.
$\Rightarrow c=-12$ ---(3).
Let us substitute equation (2) and (3) in equation (1).
So, we get ${{x}^{2}}+0x-12$.
$\Rightarrow {{x}^{2}}-12$.
So, we have got the equation of the quadratic polynomial as ${{x}^{2}}-12$.
Here we have assumed the quadratic equation as ${{x}^{2}}+bx+c$ to reduce the calculation time. We can also assume $a{{x}^{2}}+bx+c$ and use the multiplication of $\left( x-2\sqrt{3} \right)$ and $\left( x+2\sqrt{3} \right)$ to get the equation.

(ii) According to the problem, we need to find the value of k if -4 is the root of a quadratic polynomial ${{x}^{2}}-x-\left( 2k+2 \right)$.
We know that zero of the polynomial is defined as the number which makes the polynomial become zero on substituting it in the place of x.
So, we substitute -4 in place of x in the polynomial ${{x}^{2}}-x-\left( 2k+2 \right)$, which should be equal to zero.
So, we have ${{\left( -4 \right)}^{2}}-\left( -4 \right)-\left( 2k+2 \right)=0$.
$\Rightarrow 16+4-2k-2=0$.
$\Rightarrow 18=2k$.
$\Rightarrow k=\dfrac{18}{2}$.
$\Rightarrow k=9$.
So, we have found the value of k as 9.
∴ The value of k is 9.

Note: We can also use the fact that the equation of the quadratic polynomial with the zeroes a and b is $\left( x-a \right)\left( x-b \right)$ in problem (i). We should confuse the sum and product of the zeroes of quadratic polynomials. We should not make calculation mistakes while solving this problem. We can also find the value of k if the properties of zeroes of the polynomial is given in problem (ii). We can also find the sum and product of the zeroes of the polynomial in problem (ii) after finding the value of k.