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Question: If one of the zeros of the quadratic polynomial \(\left( k-1 \right){{x}^{2}}+kx+1\) is \(-3\), then...

If one of the zeros of the quadratic polynomial (k−1)x2+kx+1\left( k-1 \right){{x}^{2}}+kx+1 is −3-3, then the value of kk is
A. 43\dfrac{4}{3}
B. −43\dfrac{-4}{3}
C. 23\dfrac{2}{3}
D. −23\dfrac{-2}{3}

Explanation

Solution

First we will assume another root of the given equation as α\alpha . We have the relation between p,qp,q which are the roots of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 and the coefficients of the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 as
p+q=−bap+q=\dfrac{-b}{a}
pq=capq=\dfrac{c}{a}
From the above relation we will make an equation to find the value of kk.

Complete step-by-step solution
Given that, the quadratic equation is
(k−1)x2+kx+1=0\left( k-1 \right){{x}^{2}}+kx+1=0
Comparing the above equation with ax2+bx+c=0a{{x}^{2}}+bx+c=0, we get a=k−1a=k-1, b=kb=k, c=1c=1
Given that the value of one root is −3-3, then p=−3p=-3
Let the other root of the equation is α\alpha , then q=αq=\alpha
We know that p+q=−bap+q=\dfrac{-b}{a}
Substituting the values of p=−3p=-3, q=αq=\alpha ,b=kb=k, a=k−1a=k-1 in the above equation, we get
−3+α=−kk−1 ⇒(−3+α)(k−1)=−k ⇒−3k+3+αk−α+k=0 ⇒k(α−2)+3−α=0 ⇒k(α−2)=α−3 ⇒k=α−3α−2...(i) \begin{aligned} & -3+\alpha =\dfrac{-k}{k-1} \\\ & \Rightarrow \left( -3+\alpha \right)\left( k-1 \right)=-k \\\ & \Rightarrow -3k+3+\alpha k-\alpha +k=0 \\\ & \Rightarrow k\left( \alpha -2 \right)+3-\alpha =0 \\\ & \Rightarrow k\left( \alpha -2 \right)=\alpha -3 \\\ & \Rightarrow k=\dfrac{\alpha -3}{\alpha -2}...\left( \text{i} \right) \\\ \end{aligned}
We know that pq=capq=\dfrac{c}{a}
Substituting the values of p=−3p=-3,q=αq=\alpha ,a=k−1a=k-1,c=1c=1, we get
−3α=1k−1 ⇒α=13(1−k) \begin{aligned} & -3\alpha =\dfrac{1}{k-1} \\\ & \Rightarrow \alpha =\dfrac{1}{3\left( 1-k \right)} \\\ \end{aligned}
Substituting the value of α\alpha from the above equation in equation (i)\left( \text{i} \right), we get
k=α−3α−2 ⇒k=13(1−k)−313(1−k)−2 ⇒k=1−9(1−k)3(1−k)1−6(1−k)3(1−k) \begin{aligned} & k=\dfrac{\alpha -3}{\alpha -2} \\\ & \Rightarrow k=\dfrac{\dfrac{1}{3\left( 1-k \right)}-3}{\dfrac{1}{3\left( 1-k \right)}-2} \\\ & \Rightarrow k=\dfrac{\dfrac{1-9\left( 1-k \right)}{3\left( 1-k \right)}}{\dfrac{1-6\left( 1-k \right)}{3\left( 1-k \right)}} \\\ \end{aligned}
Now cancelling the term 3(1−k)3\left( 1-k \right) in both numerator and denominator, we get
k=1−9+9k1−6+6kk=\dfrac{1-9+9k}{1-6+6k}
Now multiplying 6k−56k-5 on both sides, we get
⇒k(6k−5)=9k−86k−5(6k−5) ⇒6k2−5k=9k−8 \begin{aligned} & \Rightarrow k\left( 6k-5 \right)=\dfrac{9k-8}{6k-5}\left( 6k-5 \right) \\\ & \Rightarrow 6{{k}^{2}}-5k=9k-8 \\\ \end{aligned}
Now adding the term 8−9k8-9k to both sides of the above equation, we get
⇒6k2−5k+8−9k=9k−8+8−9k ⇒6k2−14k+8=0 \begin{aligned} & \Rightarrow 6{{k}^{2}}-5k+8-9k=9k-8+8-9k \\\ & \Rightarrow 6{{k}^{2}}-14k+8=0 \\\ \end{aligned}
Now dividing the above equation with 22, we get
⇒12(6k2−14k+8)=0×12 ⇒3k2−7k+4=0 \begin{aligned} & \Rightarrow \dfrac{1}{2}\left( 6{{k}^{2}}-14k+8 \right)=0\times \dfrac{1}{2} \\\ & \Rightarrow 3{{k}^{2}}-7k+4=0 \\\ \end{aligned}
Solving the above equation to get the value of kk, we get
⇒3k2−3k−4k+4=0\Rightarrow 3{{k}^{2}}-3k-4k+4=0
Taking 3k3kcommon from 3k2−3k3{{k}^{2}}-3k and −4-4 from −4k+4-4k+4 , we get
⇒3k2−3k−4k+4=0 ⇒3k(k−1)−4(k−1)=0 \begin{aligned} & \Rightarrow 3{{k}^{2}}-3k-4k+4=0 \\\ & \Rightarrow 3k\left( k-1 \right)-4\left( k-1 \right)=0 \\\ \end{aligned}
Now taking (k−1)\left( k-1 \right) common from the above equation, we get
⇒(k−1)(3k−4)=0\Rightarrow \left( k-1 \right)\left( 3k-4 \right)=0
Now equating first term to zero to get the value of kk, then we will have
⇒k−1=0 ∴k=1 \begin{aligned} & \Rightarrow k-1=0 \\\ & \therefore k=1 \\\ \end{aligned}
Now equating the second term to zero to get another value of kk, then we will have
⇒3k−4=0 ⇒3k=4 ∴k=43 \begin{aligned} & \Rightarrow 3k-4=0 \\\ & \Rightarrow 3k=4 \\\ & \therefore k=\dfrac{4}{3} \\\ \end{aligned}
∴\therefore the values of kk are k=1 or 43k=1\text{ or }\dfrac{4}{3}
Hence the answer is k=1 or 43k=1\text{ or }\dfrac{4}{3}.

Note: We can also solve the above equation by simple method. i.e. substituting the zero of the equation −3-3 in the equation (k−1)x2+kx+1\left( k-1 \right){{x}^{2}}+kx+1, we get
(k−1)(−3)2+k(−3)+1=0 ⇒9(k−1)−3k+1=0 ⇒9k−9−3k+1=0 ⇒6k=8 ∴k=43 \begin{aligned} & \left( k-1 \right){{\left( -3 \right)}^{2}}+k\left( -3 \right)+1=0 \\\ & \Rightarrow 9\left( k-1 \right)-3k+1=0 \\\ & \Rightarrow 9k-9-3k+1=0 \\\ & \Rightarrow 6k=8 \\\ & \therefore k=\dfrac{4}{3} \\\ \end{aligned}
∴\therefore From both the methods we got the same answer.