Question

If the sum of the squares of zeroes of the quadratic polynomial $f\left( x \right)={{x}^{2}}-4x+k$ is 20, then find the value of k.

Answer 1

Hint: In order to find the solution of this question, we will assume the zeros or the roots of the quadratic polynomials as $\alpha$ and $\beta$. And we should remember that for any quadratic polynomial, $a{{x}^{2}}+bx+c$, the sum of roots is given by, $\dfrac{-b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$. By using this concept, we can find the answer to this question.

Complete step-by-step solution -
In this question, we have been asked to find the value of k which is in a quadratic polynomial, $f\left( x \right)={{x}^{2}}-4x+k$ whose sum of squares of zeroes is 20. To solve this question, let us consider the roots of the polynomial ${{x}^{2}}-4x+k$ as $\alpha$ and $\beta$. So, according to the question, we have been given that ${{\alpha }^{2}}+{{\beta }^{2}}=20.........\left( i \right)$.
Now, we know that the sum of roots and product of roots for any quadratic polynomial, $a{{x}^{2}}+bx+c$ is given by, $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ respectively. So, for the quadratic polynomial, ${{x}^{2}}-4x+k$, we can say that a = 1, b = -4 and c = k. Therefore, we can say that the sum of roots, that is, $\alpha +\beta =\dfrac{-\left( -4 \right)}{1}=4.........\left( ii \right)$ and the product of roots, that is, $\alpha \beta =\dfrac{k}{1}..............(iii)$.
Now, we know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. So, we can say that ${{\left( x+y \right)}^{2}}-2xy={{x}^{2}}+{{y}^{2}}$. Therefore, for values $x=\alpha$ and $y=\beta$, we get, ${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta$. Hence, we can write equation (i) as,
${{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta =20$
Now, we will take the values of $\alpha +\beta$ and $\alpha \beta$ from equations (ii) and (iii). So, we get the above equation as,
${{\left( 4 \right)}^{2}}-2k=20$
Now, we will simplify it to get the value of k. So, we get,
16 – 20 = 2k
2k = -4
$k=\dfrac{-4}{2}$
$k = -2$
Hence, we can say that the value of k is -2, in the quadratic polynomial, $f\left( x \right)={{x}^{2}}-4x+k$ whose sum of square of zeroes is 20.

Note: While solving this question, the possible mistake one can make is by finding the zeros of the polynomial $f\left( x \right)={{x}^{2}}-4x+k$ which is not wrong, but it will be a lengthy and complicated method to find the answer of this question. Also, there are chances of calculation mistakes so we have to be very careful while doing the calculations. Also, one must know that zeroes of a polynomial are also known as the roots of the polynomial.