Question

Q 13. The value of k for which the given equation kx(x - 3 ) + 6 = 0 have real and equal roots *​

Answer 1

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

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...

Sᴏʟᴜᴛɪᴏɴ :-

→ kx(x - 3) + 6 = 0

→ kx² - 3kx + 6 = 0

comparing it with A•x^2 + B•x + C = 0, we get,

• A = k
• B = (-3k)
• C = 6

since roots are real and equal , D = 0

→ B^2 - 4•A•C = 0

→ (-3k)² - 4*k*6 = 0

→ 9k² - 24k = 0

→ 9k² = 24k

→ 9k = 24

→ 3k = 8

→ k = (8/3) (Ans.)

Hence, value of k will be (8/3) .

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• kx(x-3) + 6 = 0

${\bf{\blue{\underline{To\:Find:}}}}$

• Value of k.

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{k {x}^{2} - 3kx + 6 = 0}}$

Compare it with ax² + bx + c = 0

Where a = k , b = -3k , c = 6

$: \implies{\sf{ d = {b}^{2} - 4ac}}$

$: \implies{\sf{ d = {( - 3k)}^{2} - 4(k)(6)}}$

$: \implies{\sf{ d = {9k}^{2} - 24k}}$

• Now given equation has two equal real roots b² - 4ac =0

$: \implies{\sf{ {9k}^{2} - 24k = 0}}$

$: \implies{\sf{ 3k(3 k - 8) = 0}}$

$: \implies{\sf{ 3 k - 8= 0}}$

$: \implies{\sf{ 3 k = 8}}$

$: \implies\boxed{\sf{ k = \frac{8}{3} }}$