Question

the values of k for which quadratic equation kx²-5x+k=0 has equal roots (a) 5/2 (b) -5/2 (c) ±5/2 (d) none of these​

Answer 1

Answer:

answer is (a) 5/2

EXPLANATION

p(x) -> kx² -5x +k

for equal roots D( b²-4ac) = 0

a= k, b= -5 , c= k

b²-4ac =0

(-5)²-4(k)(k)=0

25 - 4k²=0

25 = 4k²

k²= 25/4

k= 5/2

Answer 2

Answer:

Given :-

• The quadratic equation kx² - 5x + k = 0 has equal roots.

To Find :-

• What is the value of k.

Solution :-

Given Equation :

$\bigstar\: \: \bf{kx^2 - 5x + k =\: 0}$

By comparing with ax² + bx + c = 0 we get,

• a = k
• b = - 5
• c = k

$\leadsto$ The quadratic equation has equal roots.

Hence,

$\mapsto \sf\bold{\purple{D =\: 0}}$

where,

• D = Discriminant

Now, as we know that :

$\clubsuit$ Discriminant Formula :

$\mapsto \sf\boxed{\bold{\pink{Discriminant\: (D) =\: b^2 - 4ac}}}$

According to the question by using the formula we get,

$\longrightarrow \bf{D =\: 0}$

$\longrightarrow \sf (- 5)^2 - 4(k)(k) =\: 0$

$\longrightarrow \sf (- 5)(- 5) - 4 \times k \times k =\: 0$

$\longrightarrow \sf 25 - 4k^2 =\: 0$

$\longrightarrow \sf {\cancel{-}} 4k^2 =\: {\cancel{-}} 25$

$\longrightarrow \sf 4k^2 =\: 25$

$\longrightarrow \sf k^2 =\: \dfrac{25}{4}$

$\longrightarrow \sf k =\: \pm \sqrt{\dfrac{25}{4}}$

$\longrightarrow \sf\bold{\red{k =\: \pm \dfrac{5}{2}}}$

${\small{\bold{\underline{\therefore\: The\: value\: of\: k\: is\: \pm \dfrac{5}{2}\: .}}}}$

Hence, the correct options is option no (c) ± 5/2 .