Question

Find the value of k by
quardractic equation two real and equal roots
5x - 2kx +20=0​

Answer 1

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* 5* 20=0`

`\\\qquad\quad\displaystyle\sf {:}\longrightarrow 4k^2-20* 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=\frac {400}{4}`

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

`\\\qquad\quad\displaystyle\sf {:}\longrightarrow k=√(100)`

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

`\therefore\sf k=10`

Answer 2

k = +10 or -10

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 * 5 * 20 = 0`

`= > 4 {k}^(2) - 400 = 0`

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

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

`= > k = √(100) = + 10 \: or \: - 10`