Question

Given that the equation kx²-20x-25=0 has equal roots, find the value of k. Also, find the roots​

Answer 1

Answer:

the equation contains equal roots. only if

b²-4ac = 0

here, in the given equation,

a = k

b = -20

c = -25

putting all this in the equation we get,

(-20) ² - 4(k) (-25) = 0

400 +100k = 0

100K = -400

k = -400/100

k = -4

hence the required quadratic equation will be,

-4x²-20x-25 = 0

4x²+20x +25 = 0

4x²+10x+10x+25 = 0

2x ( 2x+5) +5(2x+5) = 0

hence, 2x+5 = 0 & 2x+5=0

2x= -5 2x = -5

x= -5/2 x = -5/2

this is the roots of given equation

Answer 2

Answer:As they have equal roots the answer is always goin to be 0

let us compare the equation with  ax²+bx+c=0 so we get.

a=k, b= -20, c= -25

Δ= b²-4ac

0 = (-20) ²-4(k X -25)

0 = 400 - 4(-25k)

0=400-(4 X -25k)

0 = 400 -(-100k)

0 = 400 + 100k

-100k= 400

-k = 400/100

-k = 4

∴k = -4

for finding the roots we already have  the value of k that is "a" is known now

so "a"=-4

"b" = -20

"c"=-25

b²-4ac= (-20)²-4(-4 X -25)

          = 400 - 4(100)

          = 400 - 400

          = 0

* note - no need to solve this because already in the question they told that it has equal roots.

x = -b ±$\sqrt{b^{2}-4ac \\}$/2a

   = -(-20)±0/2 X-4

   =20 / -8

   = 10/-4(divided by 2)

   = 5/-2

as they said roots are equal so 5 /-2 is the answer.