Question

4. Determine the value of m for which the
equation (4-m) x2 + (2m+4)x+(8m+1) =0
may have equal roots.​

Answer 1

Step-by-step explanation:

Given :-

(4-m)x²+(2m+4)x+(8m+1) = 0

To find :-

Determine the value of m for which the

equation (4-m) x2 + (2m+4)x+(8m+1) =0

may have equal roots.?

Solution :-

Given Quadratic equation is

(4-m)x²+(2m+4)x+(8m+1) = 0

On Comparing this with the standard quadratic equation ax²+bx+c = 0

We have,

a = 4-m

b = 2m+4

c = 8m+1

We know that

If a quadratic equation ax²+bx+c = 0 has equal roots then the discriminant (D)=b²-4ac = 0.

Now,

The discriminant (D)=b²-4ac = 0.

=>(2m+4)²-4(4-m)(8m+1) = 0

=> (2m)²+2(2m)(4)+(4)²-4(4-m)(8m+1) = 0

Since (a+b)² = a²+2ab+b²

=> 4m²+16m+16-4(32m+4-8m²-m) = 0

=> 4m²+16m+16-4(-8m²+31m+4) = 0

=>4(m²+4m+4)-4(-8m²+31m+4) = 0

=> 4[(m²+4m+4)-(-8m²+31m+4) ]=0

=> (m²+4m+4)-(-8m²+31m+4) = 0/4

=> (m²+4m+4)-(-8m²+31m+4) = 0

=> m²+4m+4+8m²-31m-4 = 0

=> (m²+8m²)+(4m-31m)+(4-4) = 0

=> 9m²-27m+0 = 0

=> 9m²-27m = 0

=> 9m (m-3) = 0

=> 9m = 0 or m-3 = 0

=> m = 0/9 or m= 0+3

=> m = 0 or m = 3

Therefore, m = 0 or 3

Answer:-

The value of m for the given problem is 0 and 3

Check:-

If m = 3 then (4-m)x²+(2m+4)x+(8m+1) = 0

=> (4-3)x²+(2(3)+4)x+(8(3)+1) = 0

=> x²+10x+25 = 0

=> x²+2(x)(5)+5² = 0

=> (x+5)² = 0

=> (x+5)(x+5) = 0

=> x = -5 and -5

And

If x = 0 then (4-m)x²+(2m+4)x+(8m+1) = 0

=> (4-0)x²+(2(0)+4)x+(8(0)+1) = 0

=> 4x²+4x+1 = 0

=> (2x)²+2(2x)(1)+1² = 0

=> (2x+1)² = 0

=> (2x+1)(2x+1) = 0

=> 2x+1 = 0

=> x = -1/2 and -1/2

Verified the given relations in the given problem.

Used formulae:-

• (a+b)² = a²+2ab+b²

• The standard quadratic equation is ax²+bx+c = 0

• If a quadratic equation ax²+bx+c = 0 has equal roots then the discriminant (D)=b²-4ac = 0.

Points to know:-

• If (D)=b²-4ac < 0 then it has no real roots.

• If (D)=b²-4ac > 0 then it has two distinct real roots.