Question

Find the value of k of the following quadratic equation so that it has two real and equal roots;
5x square 2- 2kx+20​

Answer 1

The given quadratic equation:

$5x^{2}$ - 2kx + 20 = 0

Here, a = 5, b = - 2k and c = 20

We have to find, the value of k.

Solution:

Discriminant, D = $b^{2}$ - 4ac

= $(-2k)^{2}$ - 4(5)(20)

= $4k^{2}$ - 400

If the roots of equations are real and equal, we must have:

D = 0

∴ $4k^{2}$ - 400 = 0

⇒ $4k^{2}$ - 400 = 0

⇒ $4k^{2}$ = 400

⇒ $k^{2}$ = 100

⇒ k = ± 10

∴ k = 10 or, - 10

Answer 2

Given :- Find the value of k of the following quadratic equation so that it has two real and equal roots;

5x² - 2kx + 20 = 0 .

Solution :-

we know that, If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

• D = B^2 - 4•A•C
• If D = 0 , then the given quadratic equation has real and equal roots.
• If D > 0 , then the given quadratic equation has real and distinct roots.
• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

So,

comparing the given quadratic equation 5x² - 2kx + 20 = 0 with A•x^2 + B•x + C = 0 we get,

• A = 5
• B = (-2k)
• C = 20 .

Since roots are real and equal .

• D = 0

Putting values we get,

→ B² - 4AC = 0

→ (-2k)² - 4 * 5 * 20 = 0

→ 4k² - 400 = 0

→ 4k² = 400

dividing both sides by 4,

→ k² = 100

square root both sides,

→ k = ± 10 . (Ans.)

Hence, value of k is ±10 .