Question

(1) If the roots of quadratic equation 2x2 + 6x + k = 0 are real and
equal then find the value of k.
tien​

Answer 1

Answer:

k=36

please mark me as the brainliest.

Step-by-step explanation:

from the question, the roots of quadratic equation 2x2 + 6x + k = 0 are real and

equal then , a=2. b=6. c =k

they are real and equal

therefore , b²-4ac=0

6²-4(2×k)=0

36-8k=0

36-k=0/8

36-k=0

-k=-36

therefore k = 36

Answer 2

❍ we are given that the equation$\sf 2x²+6x+k = 0$ have equal and real roots. All we have to find is the value of k.

Let's get started

___________________

We know that :-

$\: \: \: \: \: \: \: \: \bigstar{\underline {\boxed{ \pink{\frak {D(discriminant) = {b}^{2} - 4ac }} }}} \: \: \: \: \:$

• here, a,b, and c are coefficient of x², coefficient of x and constant respectively.
• In the given equation :-

$\: \: \: \: \mapsto \frak{b = 6}$

$\: \: \: \: \mapsto \frak{a= 2}$

$\: \: \: \: \mapsto \frak{c = k}$

Now, it is given if the equation has equal and real roots, then it's D(discriminant) = 0

which means that :-

$: \implies \underline {\boxed {\sf {b}^{2} - 4ac = 0}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \dag \: \underline \frak{substituing \: the \: values \: in \: equation : - }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto \bf {b}^{2} - 4ac = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto \bf {6}^{2} - 4(2)(k) = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto \bf 36 - 8k = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto \bf k = \dfrac{ \cancel{36 ^{9} }}{ \cancel8_2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto{\boxed {\pink{{\bf{ k = \frac{9}{2} }}}}} \bigstar$

I hope it's beneficial :D