Question

x
5
Find "k' if 9x² - 5 kx + 25 = 0 has real and equal
roots.​

Answer 1

Answer:

$\sf{The \ value \ of \ k \ is \ 6 \ or \ -6.}$

Given:

• $\sf{The \ given \ quadratic \ equation \ is}$$\sf{9x^{2}-5kx+25=0}$
• $\sf{\leadsto{Equation \ has \ real \ and \ equal \ roots.}}$

To find:

• $\sf{The \ value \ of \ k.}$

Solution:

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{9x^{2}-5kx+25=0}$

$\sf{Here, \ a=9, \ b=-5k \ and \ c=0}$

$\sf{Discriminant(\Delta)=b^{2}-4ac}$

$\sf{But, \ roots \ are \ real \ and \ equal.}$

$\sf{\therefore{\Delta=0}}$

$\sf{\therefore{b^{2}-4ac=0}}$

$\sf{\therefore{(-5k)^{2}-4(9)(25)=0}}$

$\sf{\therefore{25k^{2}-900=0}}$

$\sf{\therefore{25k^{2}=900}}$

$\sf{\therefore{k^{2}=\dfrac{900}{25}}}$

$\sf{\therefore{k^{2}=36}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\leadsto{k=\pm6}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 6 \ or \ -6.}}}$