Question

one zero of the Polynominal
6x^2 -px -3 is 3/2. The zeroes of the polynomial px2+x+k are equal. Then the value of K is?

a) -1/14
b) 1/28
c) 1/7
d) 7

Answer 1

Step-by-step explanation:

Given: One zero of the polynomial $6x^2 -px -3$ is 3/2. The zeroes of the polynomial $p {x}^{2} +x+k$ are equal.

To find: Value of k is:

a) -1/14

b) 1/28

c) 1/7

d) 7

Solution:

Tip:

1) Use relation of zeros and coefficient of polynomial.

2) Use Discriminant equal to zero.

Step 1: Find value of p.

Let the zeros of polynomial are $\alpha \: and \: \beta$.

We know that relation between zeros and coefficient of polynomial is given by

$\bold{\alpha + \beta = - \frac{b}{a}}$

$\bold{\alpha \beta = \frac{c}{a}}$

let

$\alpha = \frac{3}{2}$

Thus,

$\frac{3}{2} + \beta = \frac{p}{6} \: ...eq1$

$\frac{3 \beta }{2} = \frac{ - 3}{6} \\ \\ \frac{3 \beta }{2} = \frac{ - 1}{2} \\ \\ \beta = \frac{ - 1}{3}$

put value of $\beta$ in eq1

$\frac{3}{2} - \frac{1}{3} = \frac{p}{6} \\ \\ \frac{9 - 2}{6} = \frac{p}{6} \\ \\ \frac{7}{6} = \frac{p}{6}$

$\bf p = 7$

Step 2: Put $p = 7$ in second polynomial.

$7 {x}^{2} + x + k$

as roots are equal in this polynomial.

Put

$D = 0$

${b}^{2} -4 ac = 0$

or

$1 - 4(7)(k) = 0$

or

$\bf k = \frac{1}{28}$

Final answer:

$\bf k = \frac{1}{28}$

Option B is correct.

Hope it helps you.

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