Question

find the value of k for which the quadratic equation 3x^2+2x+k=0 two equal real roots​

Answer 1

AnswEr :

Given Quadratic Equation :

$\large\dag\: \sf 3x^2 + 2x + k = 0$

We've to find the value of k for equation should've two equal roots. We know the formula -

$\large\boxed{\mathfrak{D = b^2 - 4ac}}$

$\frak{Here}\begin{cases}\sf{D \: is \: Discriminant} \\ \sf{Value \: of \: a \: is \: 3}\\\sf{Value \: of \: b \: is \: 2}\\\sf{Value\: of \: c \: is \: k} \end{cases}$

$\rule{200}2$

$\large\implies\sf D = b^2 - 4ac$

$\large\implies\sf \Big(2 \Big)^2 - 4 \Big( 3 \Big) \Big( k \Big) = 0$

$\large\implies\sf 4 - 12k = 0$

$\large\implies\sf -12k = - 4$

$\large\implies\sf k = \cancel\dfrac{-4}{-12}$

$\large\implies\boxed{\sf{\red{k = \dfrac{1}{3}}}}$

For real and equal roots value of k will be 1/3.

$\rule{200}2$

Answer 2

Step-by-step explanation:

GIVEN QUADRATIC EQUATION IS↙️

3x^2+2x+k=0,and roots are real!

here,

a=3

b=2

c=k

$as \: we \: know \: that \: \: d = b ^{2} - 4ac$

$putting \: the \: value \: of \: a = 3 \: \: \: b = 2 \: \: and \: c = k$

$we \: \: get \:$

$2 ^{2} - 4 \times 2 \times k$

$4 - 12k$

the given equation will have real roots if,

$d \geqslant 0$

$4 - 12k \geqslant 0$

$12k \leqslant 4$

$k \leqslant \frac{4}{12}$

$k \leqslant \frac{1}{3}$

$<font color =green>$

$therefore \: \: th \: value \: of \: k \: \leqslant \frac{1}{3}$