Question

if one zero of a polynomial 3x2-8x+2k+1 is seven times the other then find the value of k.
Please help me it's urgent.... ​

Answer 1

Answer

The value of k is 2/3

Given

The quadratic polynomial is :

• 3x² - 8x + 2k + 1
• One zero of the given polynomial is 7 times the other

To Find

• The value of k

Solution

Let us consider one zero of the polynomial be α , so other zero is 7α

From the relationship of sum of zeroes and coefficients we have:

$\sf sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\sf \implies \alpha + 7\alpha = -\dfrac{-8}{3} \\\\ \sf\implies 8\alpha = \dfrac{8}{3} \\\\ \sf\implies \alpha = \dfrac{1}{3}$

Now from product relation zeroes and coefficients :

$\sf product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ \sf \alpha \times 7 \alpha = \dfrac{2k + 1}{3} \\ \\ \sf \implies 7 \alpha {}^{2} = \dfrac{2k + 1}{3} \\ \\ \sf \implies 21( \dfrac{1}{3} ) {}^{2} = 2k + 1 \\ \\ \sf \implies2k + 1 = 21 \times \dfrac{1}{3 \times 3} \\ \\ \sf \implies 2k = \dfrac{7}{3} - 1 \\ \\ \sf\implies k = \frac{7 - 3}{3 \times 2} \\ \\ \sf \implies k = \dfrac{4}{6} \\ \\ \sf \implies k = \dfrac{2}{3}$

Thus , the value of k is 2/3

Answer 2

Answer:

2/3

Step-by-step explanation:

Assume that the one zero is x. So, the other zero is 7x.

Sum of zeros = -b/a

→ x + 7x = -(-8)/3

→ 8x = 8/3

→ x = 1/3

Product of zeros = c/a

→ x(7x) = (2k + 1)/3

→ 7x² = (2k + 1)/3 ..........(1)

Substitute value of x = 1/3 in (1)

→ 7(1/3)² = (2k + 1)/3

→ 7/3 = 2k + 1

→ 7 = 6k + 3

→ 6k = 4

→ k = 4/6

→ k = 2/3

Hence, the value of k is 2/3.