Question

Q.1 If one of the zeroes of the quadratic polynomial (k-1) x² + kx + 1 is -3, then the value of k is.
(2) 4/3
(b) -4/3
(c) 2/3
(d)-2/3​

Answer 1

Answer:

$\large\bold\red{(a)\frac{4}{3}}$

Step-by-step explanation:

Given,

a quadratic Polynomial,

$(k - 1) {x}^{2} + kx + 1$

Also,

One of its root is -3

Let the another root be 'p'

Now we know that,

General form of quadratic polynomial is

$a {x}^{2} + bx + c$

Therefore,

Comparing the coefficients,

we get,

$a = (k - 1) \\ \\ b = k \\ \\ c = 1$

Now,

we know that,

Product of roots of quadratic equation is $\bold{\frac{c}{a}}$

$= > - 3p = \frac{1}{k - 1} \\ \\ = > 1 - k = \frac{1}{3p} \\ \\ = > k = 1 - \frac{1}{3p} \: \: \: \: \: ................(i)$

Also,

sum of roots of quadratic equation is $\bold{\frac{-b}{a}}$

$= > - 3 + p = \frac{ - k}{k - 1}$

Putting the value of 'p'

we get,

$= > - 3 + \frac{1}{3(1 - k)} = \frac{k}{1 - k} \\ \\ = > \frac{1}{1 - k} ( \frac{1}{3} - 3(1 - k) - k) = 0 \\ \\ = > \frac{1}{3} - 3 + 3k - k = 0 \\ \\ = > 2k = 3 - \frac{1}{3} \\ \\ = > 2k = \frac{8}{3} \\ \\ = > k = \frac{8}{3 \times 2} \\ \\ = > k = \frac{4}{3}$

Hence,

Correct Option is $\bold{(a)\frac{4}{3}}$