If the sum and product of the roots of the equation k * x ^ 2 + 6x + 4k = 0 are equal then k =

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If the sum and product of the roots of the equation kx2 + 6x + 4k = 0 are real, then k =

Question:

If the sum and product of the roots of the equation $k x^{2}+6 x+4 k=0$ are real, then $k=$

(a) $-\frac{3}{2}$

(b) $\frac{3}{2}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

Solution:

The given quadric equation is $k x^{2}+6 x+4 k=0$, and roots are equal

Then find the value of $c$

Let $\alpha$ and $\beta$ be two roots of given equation

And, $a=k, b=6$ and,$c=4 k$

Then, as we know that sum of the roots

$\alpha+\beta=\frac{-b}{a}$

$\alpha+\beta=\frac{-6}{k}$

And the product of the roots

$\alpha \cdot \beta=\frac{c}{a}$

$\alpha \beta=\frac{4 k}{k}$

$=4$

According to question, sum of the roots = product of the roots

$\frac{-6}{k}=4$

$4 k=-6$

$k=\frac{-6}{4}$

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

Therefore, the value of $c=\frac{-3}{2}$

Thus, the correct answer is $(a)$

The given quadric equation is kx2 + 6x + 4k = 0, and roots are equal

Then find the value of c.

Let `alpha and beta`be two roots of given equation

And,  a = k,b = 6 and , c = 4k 

Then, as we know that sum of the roots

`alpha + beta = (-b)/a`

`alpha +beta = (-6)/a`

And the product of the roots

`alpha. beta = c/a`

  `alpha beta = (4k)/k`

        = 4

According to question, sum of the roots = product of the roots

`(-6)/k = 4`

    `4k = -6`

      `k = (-6)/4`

         `= (-3)/2`

Therefore, the value of `c = (-3)/2`.

Page 2

The given quadric equation is `ax^2 + bx + c = 0`, and `sin alpha and cos beta` are roots of given equation.

And, a = a,b = b and, c = c

Then, as we know that sum of the roots

 `sin alpha + cos beta - (-b)/a`…. (1)

And the product of the roots

`sin alpha .cos beta =c/a`…. (2)

Squaring both sides of equation (1) we get

`(sin alpha + cos beta)^2 = ((-b)/a)^2`

`sin^2 alpha + cos^2 beta + 2 sin alpha cos beta = b^2/a^2`

Putting the value of `sin^2 alpha + cos^2 beta = 1`, we get

`1 + 2 sin alpha cos beta = b^2/a^2`

`a^2 (1+2 sin alpha cos beta) = b^2`

Putting the value of`sin alpha.cos beta = c/a` , we get

`a^2 (1 + 2 c/a) = b^2`

`a^2 ((a+2c)/a) = b^2`

        `a^2 + 2ac =b^2`

Therefore, the value of `b^2 = a^2 + 2ac`.

If the sum and product of the roots of the equation k x2+6 x+4 k=0 are real, then k=a 32b 32c 23d 23