Choose the correct answer from the given four options in the following questions: If one of the zeroes of the quadratic polynomial (k-1) x² + k x + 1 is –3, then the value of k is 4. A 4/3 (B) -4/3 (C) ⅔ (D)-⅔
2 thoughts on “Choose the correct answer from the given four options in the following questions:<br /> If one of the zeroes of the quadratic poly”
ANSWER:
Given:
p(x) = (k – 1)x² + kx + 1
-3 is a zero of p(x)
To Find:
Value of k
Solution:
[tex]\text{We are given that,}\\\\:\longrightarrow p(x)=(k-1)x^2+kx+1\\\\\text{We are also given that, -3 is a zero of p(x).}\\\\\text{So,}\\\\:\implies p(-3)=0\\\\\text{Hence,}\\\\:\implies p(-3)=(k-1)(-3)^2+k(-3)+1=0[/tex]
[tex]:\implies9(k-1)-3k+1=0\\\\:\implies9k-9-3k+1=0\\\\:\implies9k-3k-9+1=0\\\\:\implies6k-8=0\\\\:\implies6k=8\\\\:\implies k=\dfrac{8\!\!\!/^{\:4}}{6\!\!\!/_{\:3}}\\\\\bf{:\implies k=\dfrac{4}{3}}\\\\\text{\bf{Hence, option A) $\dfrac{4}{3}$ is the correct value of k.}}[/tex]
ANSWER:
Given:
To Find:
Solution:
[tex]\text{We are given that,}\\\\:\longrightarrow p(x)=(k-1)x^2+kx+1\\\\\text{We are also given that, -3 is a zero of p(x).}\\\\\text{So,}\\\\:\implies p(-3)=0\\\\\text{Hence,}\\\\:\implies p(-3)=(k-1)(-3)^2+k(-3)+1=0[/tex]
[tex]:\implies9(k-1)-3k+1=0\\\\:\implies9k-9-3k+1=0\\\\:\implies9k-3k-9+1=0\\\\:\implies6k-8=0\\\\:\implies6k=8\\\\:\implies k=\dfrac{8\!\!\!/^{\:4}}{6\!\!\!/_{\:3}}\\\\\bf{:\implies k=\dfrac{4}{3}}\\\\\text{\bf{Hence, option A) $\dfrac{4}{3}$ is the correct value of k.}}[/tex]
Answer:
(A) 4/3
Step-by-step explanation:
-3 is zero of (k-1) x^2 +kx+1
now, (k-1) (-3) ^2 +k(-3) +1 =0
(k-1) (9) -3k+1=0
9k-9-3k+1 = 0
6k -8 = 0
6k = 8
k= 8/6
k=4/3