Question

Find the value of k byquardractic equation two real and equal roots5x - 2kx +20=0​

Answer 1

Step-by-step explanation:

Given Equation

5x-2kx+20=0

• If it has real and equal roots then

$\boxed{\sf \longrightarrow D=0 }$

• Substitute the values

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow b^2-4ac=0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow (-2k)^2-4\times 5\times 20=0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 4k^2-20\times 20=0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 4k^2-400=0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 4k^2=400$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow k^2=\dfrac {400}{4}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow k^2=100$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow k=\sqrt{100}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow k=10$

$\therefore\sf k=10$

Answer 2

Answer:

k = +10 or -10

Step-by-step explanation:

It's given in the question that the roots of the eqn. are real and equal. So , the discriminant of the eqn. should be equal to 0.

$= > {b}^{2} - 4ac = 0$

$= > {( - 2k)}^{2} - 4 \times 5 \times 20 = 0$

$= > 4 {k}^{2} - 400 = 0$

$= > 4( {k}^{2} - 100) = 0$

$= > {k}^{2} - 100 = 0$

$= > k = \sqrt{100} = + 10 \: or \: - 10$