Question

It is known that x1 and x2 are roots of the equation 6x2+7x+k=0, where 2x1+3x2=−4.
Find k.

Answer 1

k=-5

Explanation:

6x^2+7x+k=0 is a quadratic equation.

a= 6; b=7; c=k

Let the roots of the equation be R1 and R2

R1+R2 = -b/a = -7/6 ---------1

R1xR2 = c/a =k/6 or (R1xR2)6=k------------2

From Equation 1:

R1=-7/6-R2

We know 2R1+3R2=-4 Substituting for R1, we get

3 (-7/6-R2)+3R2=-4

R2=-5/3

R1=-7/6-(-5/3)= 1/2

Substituting these values in Eq2,

k= (-5/3 x 1/2) 6

k=-5

Answer 2

1

Explanation:

For a quadratic equation, the roots are expressed by the quadratic formula.

x=(-b+/- Sqrt[b^2-4ac])/2a

In this case a=6, b=-7 and c=k

So,

x=(7 +/- √[(-7)^2-4(6)(k)]/2(6))

Simplifying gives:

x=(7 +/- √[49-24k])/12

For k=0 the square root simplifies to √[49]=7 which yields roots of 7/6 and 0

For k=1 the square root simplifies to √[49-24]=√[25]=5 which yields roots of 1 and 1/6

For k=2 the square root simplifies to √[49-48]=√[1]=1 which yields roots of 2/3 and 1/2

k= 1 as other roots are fractions