Question

Find the value of 'm' , if the roots of the following quadratic equation (4+m)x² + (m+1)x+1=0


❤️PLEASE HELP ❤️

Answer 1

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HEY THERE ,

For , the given equation (4+m)x² + (m+1)x+1=0;

a=4+m, b= m+1 and c=1.

Since , the roots are equal

Therefore,

b² - 4ac = 0

=> (m+1)² -4(4+m)×1=0

=>m²+2m+1-16-4m=0

=>m²-2m-15=0

ON SOLVING , WE GET,

m=5 , m=-3.

Thanks.

Answer 2

Answer:

The roots of the equation are equal .

The given equation is ( 4 + m ) x² + ( m + 1 ) x + 1 = 0 .

Comparing with a x² + bx + c = 0 :

a = 4 + m

b = m + 1

c = 1

When the roots of the equation are equal , then we can write that b² = 4 ac  .

Hence :

( m + 1 )² = 4 ( 4 + m ) 1

⇒ m² + 1 + 2 m = 16 + 4 m

⇒ m² - 2 m - 15 = 0

Splitting - 2m into 3 m - 5 m we get :-

⇒ m² + 3 m - 5 m - 15 = 0

Take commons :-

⇒ m ( m + 3 ) - 5 ( m + 3 ) = 0

⇒ ( m - 5 )( m + 3 ) = 0

Either m = 5 .

Or m = - 3

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Step-by-step explanation:

It is not mentioned in the question .

The roots of the equation will be equal .

When roots are equal :

b² = 4 ac

When roots are unequal and real :-

b² > 4 ac

When roots are complex :

b² < 4 ac

Apply the above formula and then find the value of m :) .