# 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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In mathematics, a linear equation is an equation that may be put in the form {\displaystyle a_{1}x_{1}+\ldots +a_{n}x_{n}+b=0, } where x_{1}, \ldots, ...

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 betabe 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 ofsin 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