Question

For what value of k the roots of the quadratic equation 3 X square + 2 k x Plus 27 real and equal

Answer 1

AnswEr:

Given Equation -

• 3x² + 2kx + 27 = 0

$\sf{Here}\begin{cases}\sf{a\:=\;3}\\ \sf{b\:=\; 2k}\\ \sf{c\;=\; 27}\end{cases}$

The Roots of the Quadratic Equation are real and Equal. [Given]

$\rule{150}2$

D = b² - 4ac

= (2k)² - 4 × 3 × 27 = 0

= 4k² - 324 = 0

= k² = $\sf\cancel\dfrac{324}{4}$

= k² = 81

= k = √81

= k = +9

Hence, Value of k is 9 & -9.

$\rule{150}2$

Additional Information -

If b² - 4ac > 0

So, The roots of the Quadratic Equation are real & unequal.

If b² - 4ac = 0

So, The roots of the Quadratic Equation are real & equal.

If b² - 4ac < 0

So, The roots of the Quadratic Equation are Imaginary.

Answer 2

☯ GiveN :

↪ Equation : 3x² + 2kx + 27

And, also given that equation has real amd equal roots.

So, D = 0

$\rule{200}{1}$

☯ To FinD :

We have to find the value of k for which equation has real and equal roots.

$\rule{200}{1}$

☯ SolutioN :

We know the discriminant method and in that D will be 0.

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

Where,

• a = 3
• b = 2k
• c = 27

★ Putting Values ★

As, we know k can't be negative.

So, k = 9

$\rule{200}{2}$

☯ VerificatioN :

Fir verification put value of k and put it equal to the 0.

$\sf{\dashrightarrow 0 = (2k)^2 - 4(3)(27)} \\ \\ \sf{\dashrightarrow 0 = 4(9)^2 - 12 \times 27} \\ \\ \sf{\dashrightarrow 0 = 4(81) - 324} \\ \\ \sf{\dashrightarrow 0 = 324 - 324} \\ \\ \sf{\dashrightarrow 0 = 0} \\ \\ \bf{Hence \: Verified}$