Question

Find the value of K for the foll qudratic
equation so that is has two real and equal
roots,
5x²-2kx+20=0​

Answer 1

$\huge{\underline{\underline{\boxed{\sf{\purple{Answer}}}}}}$

5x²-2kx+20 = 0

a = 5

b = -2k

c = 20

Discriminent (d) = b² - 4ac

For equal roots :

b² - 4ac = 0

(-2k)² - 4(5)(20) = 0

4k² - 400 = 0

4k² = 400

k² = 400/4

k² = 100

k = √100

k = 10

Answer 2

Given :- 5x²-2kx+20=0

To Find :- Value of K?

→ 5x²-2kx+20=0

Here,

→ a = 5

→ b = -2k

→ c = 20.

We know that :-

• Discriminent b = b² -4ac

For the equal roots :-

→ b² -4ac = 0

→ (-2k)² -4(5)(20) = 0

→ 4k² -400 = 0

→ 4k² = 400

→ k² = 400/4

→ k² = 100

→ k = √100

→ k = 10

So, the value of k is 10.