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Question: If the squared difference of the zeroes of the polynomial \(p\left( x \right)={{x}^{2}}+3x+k\) is 3,...

If the squared difference of the zeroes of the polynomial p(x)=x2+3x+kp\left( x \right)={{x}^{2}}+3x+k is 3, then find the value of k.

Explanation

Solution

We start solving the problem by recalling the definition of zeroes of a given polynomial. We then assume the variables for the zeroes of the polynomial p(x)=x2+3x+kp\left( x \right)={{x}^{2}}+3x+k. When this polynomial is equated to 0, we can see that it resembles the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0. We use the sum and product of the roots of the quadratic equation for the given polynomial and we make subsequent calculations to get the value of k.

Complete step-by-step answer :
According to the problem, we have given a polynomial p(x)=x2+3x+kp\left( x \right)={{x}^{2}}+3x+k whose squared difference of the zeroes is 3. We need to find the value of k.
We know zero of a polynomial p(x)p\left( x \right) is defined as the number r which holds the property p(r)=0p\left( r \right)=0.
Let us assume ‘s’ and ‘t’ be the zeroes of the polynomial p(x)=x2+3x+kp\left( x \right)={{x}^{2}}+3x+k.
Let us consider p(r)=0p\left( r \right)=0. So, we have r2+3r+k=0{{r}^{2}}+3r+k=0---(1).
We can see that this resembles the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0. We know that the sum and product of the zeroes of this equation is −ba\dfrac{-b}{a} and ca\dfrac{c}{a}. We use these results for equation (1).
So, we have s+t=−31s+t=\dfrac{-3}{1}.
⇒s+t=−3\Rightarrow s+t=-3 ---(2).
We also have st=k1st=\dfrac{k}{1}.
⇒st=k\Rightarrow st=k ---(3).
According to the problems, we have squared difference of the zeroes as 3.
So, we have (s−t)2=3{{\left( s-t \right)}^{2}}=3.
⇒s2+t2−2st=3\Rightarrow {{s}^{2}}+{{t}^{2}}-2st=3.
⇒s2+t2+2st−4st=3\Rightarrow {{s}^{2}}+{{t}^{2}}+2st-4st=3.
⇒(s+t)2−4st=3\Rightarrow {{\left( s+t \right)}^{2}}-4st=3.
From equations (2) and (3).
⇒(−3)2−4k=3\Rightarrow {{\left( -3 \right)}^{2}}-4k=3.
⇒9−4k=3\Rightarrow 9-4k=3.
⇒4k=9−3\Rightarrow 4k=9-3.
⇒4k=6\Rightarrow 4k=6.
⇒k=64\Rightarrow k=\dfrac{6}{4}.
⇒k=32\Rightarrow k=\dfrac{3}{2}.
We have found the value of k as 32\dfrac{3}{2}.
∴ The value of k is 32\dfrac{3}{2}.

Note : We should not confuse the sum and product of the zeroes of the quadratic equation. We should not make any calculation mistakes while solving the problem. We can verify by substituting the obtained value of k and finding the zeros of the polynomial. Similarly, we can expect problems to find the roots of the polynomial, value of the polynomial for a given value of x after finding the value of k.