Question

If a and b are the zeroes of the polynomial $f\left( x \right)={{x}^{2}}-5x+k$ such that $a-b=1$, then find the value of k.

Answer 1

Hint: To find the value of ‘k’, use the fact that the sum and the product of roots of quadratic equation of the form $p{{x}^{2}}+qx+r=0$ is given by $\dfrac{-q}{p}$ and $\dfrac{r}{p}$ respectively. Add the equation $a-b=1$ to equation formed by taking the sum of roots of the equation $f\left( x \right)={{x}^{2}}-5x+k$ to find the roots of the equation. Multiply the roots of the equation to find the value of ‘k’.

Complete step-by-step answer:
We have the polynomial $f\left( x \right)={{x}^{2}}-5x+k$ whose roots are ‘a’ and ‘b’ such that $a-b=1$. We have to find the value of ‘k’.
We know that the sum and the product of roots of quadratic equation of the form $p{{x}^{2}}+qx+r=0$ is given by $\dfrac{-q}{p}$ and $\dfrac{r}{p}$ respectively. Substituting $p=1,q=-5,r=k$ in the above expression, we observe that sum of roots is given by $-\left( \dfrac{-5}{1} \right)=5$ and product of roots is given by $\dfrac{k}{1}=k$.
Thus, we have $a+b=5.....\left( 1 \right)$ and $ab=k$.
We know that $a-b=1.....\left( 2 \right)$.
Adding equation (1) and (2), we have $\left( a+b \right)+\left( a-b \right)=5+1$. Thus, we have $2a=6$.
Rearranging the terms of the above equation, we have $a=\dfrac{6}{2}=3.....\left( 3 \right)$.
Substituting equation (3) in equation (1), we have $3+b=5$. Rearranging the terms, we have $b=5-3=2$.
Thus, we have $a=3,b=2$. Substituting these values in the equation $ab=k$, we have $2\times 3=k=6$.
Hence, the value of ‘k’ is 6.

Note: We can check if the calculated value of ‘k’ is correct or not by substituting the value of ‘k’ in the polynomial equation and factoring it to find its roots and check if it satisfies the conditions given in the question or not. It’s necessary to use the relation between coefficients of quadratic equation and sum and product of the roots; otherwise, we won’t be able to solve this question.