Question

find 'm' so that one root of the equation x^2 + (2m+1)x+ (m^2+2)=0 is twice the other​

Answer 1

Step-by-step explanation:

Given :-

One of the roots of the equation x²+ (2m+1)x+ (m²+2)=0 is twice the other root.

To find :-

Find the value of m ?

Solution :-

Given equation is x²+ (2m+1)x+ (m²+2)=0

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

a = 1

b = 2m+1

c = m²+1

Let the other root be X

The one of the roots = Twice the other = 2X

We know that

Sum of the roots = -b/a

=> X+2X = -(2m+1)/1

=> 3X = -(2m+1)

=> X = -(2m+1)/3 -----------(1)

Product of the roots = c/a

=> X×2X = (m²+1)/1

=> 2X² = m²+1

=>2[-(2m+1)]² = m²+1 (from (1))

=> 2[(2m)²+2(2m)(1)+1²] = m²+1

=> 2[4m²+4m+1] = m²+1

=> 8m²+8m+2 = m²+1

=> 8m²+8m+2-m²-1 = 0

=>7m²+8m+1 = 0

=> 7m²+7m+m+1 = 0

=> 7m(m+1)+1(m+1) = 0

=> (m+1)(7m+1) = 0

=> m+1 = 0 or 7m+1 = 0

=> m = -1 or 7m = -1

=> m = -1 or m = -1/7

Therefore, m = -1 or -1/7

Answer:-

The value of m for the given problem is -1 or -1/7

Used formulae:-

• The standard quadratic equation is ax²+bx+c = 0
• Sum of the roots = -b/a
• Product of the roots = c/a