Question

let us calculate and write the value of k for which 2by3 will be A root the quadratic equation 7xsquare+kx-3​

Answer 1

Answer: k = -1/6

Given quadratic equation, 7x² + kx - 3 = 0 ; It is given that one of the two zeroes of this quadratic equation is 2/3.

These are just assumptions, with that, we have to find the value of k.

We know, zero of a quadratic equation is that value that satisfies the quadratic equation.

Which means,

x = 2/3 then,

⇒ 7(2/3)² + k(2/3) - 3 = 0

⇒ 7×4/9 + 2k/3 - 3 = 0

⇒ (28 + 6k - 27)/9 = 0

⇒ 1 + 6k = 0

⇒ 6k = -1

⇒ k = -1/6

Hence, The value of k is -1/6 for the given conditions.

Some Information :-

• A quadratic equation has two zeroes which can be real and distinct, imaginary, real and equal. This depends upon the discriminant of the quadratic equation which is given by,

⇒ D = b² - 4ac, where

• ax² + bx + c = 0 is the quadratic equation.

Also, if

D > 0

• Real and distinct roots exist.

D = 0

• Real and equal roots exist.

D < 0

• Imaginary roots.

Answer 2

Given :-

Equation :-

$\bf \: 7 {x}^{2} + kx - 3$

$\bf \: x = \dfrac{2}{3}$

To Find :-

Value of k

Solution :-

We need to just put the value of x to find k

$\sf \: 7 {x}^{2} - kx - 3 = 0$

$\sf \: 7 \bigg( \dfrac{2}{3} \bigg)^2- k \bigg( \dfrac{2}{3} \bigg) - 3 = 0$

$\sf \: 7 \times \dfrac{2}{3} \times \dfrac{2}{3} - k \times \dfrac{2}{3} - 3 = 0$

$\sf \: 7 \times \dfrac{4}{9} - \dfrac{2k}{3} - 3 = 0$

$\sf \: \dfrac{7 \times 4 - 3(2k) - 3 \times 9}{9} = 0$

$\sf \: \dfrac{28 - 6k - 27}{3} = 0$

$\sf \dfrac{1 - 6k}{3} = 0$

$\sf \: 1 - 6k = 3 \times 0$

$\sf 6k = 0 - 1$

$\sf 6k = - 1$

k = -1/6