Question

One zero of the polynomial 6x 2 -px -3 is 3/2. The zeroes of the polynomial px2+x+k are equal . The value of k is

a. -1/14
b. 1/28
c. 1/7
d. 7​

Answer 1

Solution :

One zero of the polynomial, 6x² - px - 3 is 3/2 .

Let this polynomial be q(x)

q(x) = 6x² - px - 3

p(3/2) = 0

>> 6 × (9/4) - (3/2)p - 3 = 0

>> 27/2 - 3/2p = 3

>> 3/2p = 27/2 - 3

>> 3/2 p = 21/2

>> 3p = 21

>> p = 7

The zeroes of the second polynomial are equal .

>> 7x² + x + k

m(x) = 7x² + x + k

Let both the zeroes be l

2l = (-b/a) = -⅐ (1)

l² = (c/a) = (k/7) (2)

Divide (2) by (1)

>> l/2 = (k/7) × -7

>> l/2 = -k

>> l = -2k

m(l) = 0

>> m(-2k) = 0

>> 7(-2k)² + (-2k) + k = 0

>> 28k² - k = 0

>> k(28k-1) = 0

Either k = 0 or k = 1/28

Hence the value of k is 1/28.

The correct answer is (b)

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Answer 2

Answer:

The value of k is $\frac{1}{28}$.

Step-by-step explanation:

Given a polynomial $6x^2-px-3=0$.

Also $\frac{3}{2}$ is the zero of the polynomial.

Let us denote the polynomial $P(x)=6x^2-px-3=0$.

If a is the zero of a polynomial P(x), then P(a)=0.

Therefore, $P(\frac{3}{2})=6(\frac{3}{2})^2-p(\frac{3}{2})-3=0$.

$\frac{3p}{2}=\frac{27}{2}-3$

$\frac{3p}{2}=\frac{21}{2}$

$p=7$

Therefore, the value of $p$ is $7$.

Now we have given another polynomial $px^2+x+k=0$.

Given that, the zeroes of the polynomial are equal.

Now substituting the value of p in the above polynomial.

$7x^2+x+k=0$

Let both the zeroes of the polynomial be a.

Sum of the zeroes, $2a=-\frac{1}{7}$.

$a=-\frac{1}{14}$

Now substitute the value of a in the equation.

$7(-\frac{1}{14})^2+(-\frac{1}{14})+k=0$

$k=\frac{1}{14}-\frac{1}{28}$

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

Therefore, the value of $k$ is $\frac{1}{28}$.

Option b) is the correct answer.