Question

if the quadratic equation mx²+2root5x+15=0,has two equals roots,then find the value of m.​

Answer 1

Answer:

m = 1/3

Step-by-step explanation:

Given :

the quadratic equation mx²+2√5x+15 = 0,has two equals roots

To find :

the value of m

Solution :

The nature of roots is determined by the value of the "discriminant"

For the quadratic equation ax² + bx + c = 0 ;

discriminant is given by, D = b² - 4ac

1. If D > 0 ; the quadratic equation has two different real roots
2. If D = 0 ; the quadratic equation has two equal real roots
3. If D < 0 ; the quadratic equation has no real roots i.e., complex roots

Given quadratic equation mx²+2√5x+15 = 0,

    a = m , b = 2√5 , c = 15

Since it's given the quadratic equation has two equal roots,

⇒ D = 0

Substitute the values in discriminant formula,

   $\sf D=(2\sqrt{5})^2 -4(m)(15) =0\\\\ \sf 4(\sqrt{5})^2-60m=0 \\\\ 4(5)=60m \\\\ 20=60m \\\\ m=20/60 \\\\ m=1/3$

The value of m is 1/3

Answer 2

$\huge\sf\underline\purple{Solution}$

Given that quadratic equation has two roots Then

$\sf\orange{b²-4ac=0}$

Given QE mx²+2√5x+15 = 0,

a = m b= +2√5 c=15

So,

b²-4ac=0

(2√5 )²-4(m)(15)=0

20 - 60m =0

20 = 60m

m = 1/3

So, value of m is 1/3

$\sf\red{Know more!!}$

If $\sf\pink{b²-4ac>0 roots are real and distinct}$

If $\sf\blue{b²-4ac<0 roots are complex and conjugate}$