The positive value of m for which the roots of equation 12x^2+mx+5=0 are in the ratio 3:2 is.

Question

The positive value of m for which the roots of equation 12x²+mx+5=0 are in ratio 3:2 is:

Answer 1

12x²+mx+5=0

the roots are in the ratio 3:2, so the roots are 3x and 2x

the sum of the roots is -b/a

3x+2x=-m/12

5x=-m/12

the product of the roots is c/a

(3x)(2x)= 6x²

6x²=5/12

5x=-m/12

60x=-m

x=-m/60

6(-m/60)²=5/12

6(m²/3600)=5/12

m²/600=5/12

m²=(600)(5/12)

m²=50*5

m²=250

m=√250

m=5√10

using the Pythagorean Theorem you get...

(-5√10±√10)24

-6√10/24 and -4√10/24

-√10/4 and -√10/6 are the roots which are in the ratio of 3:2 

 

Answer 2

r₁/r₂ = 3/2

r₁ = 3r₂/2

r₁r₂ = 3(r₂)²/2 = 5/12 (product of the roots = constant term)

r₂ = (10^.5)/6 and then r₁ = (10^.5)/4

 

The coefficient in the x term of (x - r₁)(x - r₂) = m/12 = -5(10^.5)/12 

and m = -5(10^.5)