Question

find the value of k for which the roots are real and equal 9 x square - 24 x + K equal to zero​

Answer 1

Answer :-

→ K = 16

To Find:-

→ Value of K .

Explanation :-

We have ;

→ 9x² - 24x + k = 0

According to the question its roots are real and equal .

We know that if roots of any quadratic polynomial are equal so it's discriminate will zero.

If we have a quadratic equation

→ ax² + bx + c = 0. eq.(1)

then , it's discriminate D is

→ D = b² - 4ac

Compare the given equation with eq.(1)

→ a = 9 , b = -24 and c = k

therefore , it's discriminate is

$\implies \sf{ D = {b}^{2} - 4ac = 0} \\ \\ \implies \sf{ {(-24)}^{2} - 4 \times 9 \times k = 0} \\ \\ \implies \sf{576 - 36k = 0} \\ \\ \implies \sf{ 576 = 36k \: } \\ \\ \implies \sf{ k = \frac{576}{36} } \\ \\ \implies \boxed{\sf{k = 16}}$

hence K = 16 .

Now we have our quadratic equation .

→ 9x² - 24x + 16 = 0

Its discriminate is 0 ,so its roots are real and equal.

Answer 2

$\tt{\huge{\orange { Hello!!! }}} S.D$

QUESTION :

find the value of k for which the roots are real and equal 9 x square - 24 x + K equal to zero .

SOLUTION :

Given Equation :

f ( x ) = 9 X ^2 - 24 X + K = 0

All the roots are real and are equal ...

Hence ,

D = 0

B ^ 2 - 4 A C = 0

=> B ^ 2 = 4 A C

{ - 24 } ^ 2 = 36 K

=> K = { 24 × 24 } / { 6 × 6 ]

=> K = 4 × 4

=> K = 16.

So, the value of K is 16 for the above condition....

Answer : K = 16...... √√√√√√√√√√√√√√√√