Question

If the sum of roots of the equation x2 - (k + 6 ) x + 2 (2k-1) = 0
is equal to half of their product, then find the value of k.​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ x² - (k + 6)x + 2(2k - 1) = 0.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -[-(k + 6)]/1 = (k + 6).

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = 2(2k - 1)/1 = 2(2k - 1).

Sum of the roots of the equation is equal to half of their products.

⇒ (α + β) = 1/2 x (αβ).

⇒ (k + 6) = 1/2 x [2(2k - 1)].

⇒ 2(k + 6) = 2(2k - 1).

⇒ 2k + 12 = 4k - 2.

⇒ 2k + 12 - 4k + 2 = 0.

⇒ - 2k + 14 = 0.

⇒ 2k = 14.

⇒ k = 7.

                                                                                                                           

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Solution :-

$\bf x^2 - (k+6)x+2(2k-1)$

Here

a = -k + 6

b = 1

c =2(2k + 1)

Sum of zeroes

$\sf \alpha + \beta =\dfrac{-(-k+6)}1$

Now

$\sf \alpha +\beta = \dfrac{k+6}1$

$\sf \alpha +\beta = k+6$

Product of zeroes

$\alpha \beta = \sf \dfrac{2(2k+1)}{1}$

$\sf \alpha \beta =2(2k+1)$

Now

According to the question

$\sf k + 6 =\dfrac{2(2k-1)}{2}$

$\sf k + 6 = 2k + 1$

$\sf 2k - k = 6+1$

$\sf k=7$