Question

If one root of the quadratic equation x
${2}^{2}$
26x + k
is double the other, find k. Also find its
roots.​

Answer 1

k = 676/9

Roots are -13/3 , -26/3

Step-by-step explanation:

Given:

(2x² + 26x² + k) have roots.

one is double on another.

To Find:

i) The value of k

ii) Roots

How to Solve?

1. Assume a variable as roots,

2. Use Formula of Quadratic Equation,

3. Simplify = Answer.

Solution:

Comparing The given equation from ax² + bx + c I get, a = 2 , b = 26 , c = k .

Let, one root be e So, another root is 2e.

Sum of roots = - b/a

Or, (e + 2e) = - 26/2

Or, 3e = - 13

Or, e = - 13/3

Or, 2e = -26/3

$\therefore$ -13/3 and -26/3 are the roots of given equation.

Now,

Product of roots = c/a

Or, 2e × e = k/2

Or, 2e² × 2 = k

Or, 4 × (-13/3)² = k

Or, 4 × 169/9 = k

Or, k = 676/9

$\therefore$The required value of k = 676/9

Formula of Quadratic equation

$</u>\textsf{Sum of roots}= \dfrac{- b}{a} \\ \\ \textsf{Product of roots} = \dfrac{c}{a}\\ \\</p><p> \textsf{Roots}= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[When equation is ax² + bx + c = 0]