Asked by Joe
Determine the range of the values of k at which the quadratic equation 3x^2+8x+2k=0 will have 2 different negative real number solutions.
Answers
Answered by
oobleck
the discriminant is 64-24 = 40
since it is positive, there will be two real roots for all values of k.
Since we want negative roots, k must be positive.
since the roots are different, consider
3(x^2 + 8/3 x + 16/9) + 2k - 3*16/9 = 0
3(x^2 + 4/3)^2 + 2k-16/3 = 0
so 2k > 16/3
k > 8/3
check:
k=2: 3x^2+8x+4 = (3x+2)(x+2) has two different negative roots
k=8/3: a repeated root at x = -4/3
k=3: 3x^2+8x+6 has no real roots
since it is positive, there will be two real roots for all values of k.
Since we want negative roots, k must be positive.
since the roots are different, consider
3(x^2 + 8/3 x + 16/9) + 2k - 3*16/9 = 0
3(x^2 + 4/3)^2 + 2k-16/3 = 0
so 2k > 16/3
k > 8/3
check:
k=2: 3x^2+8x+4 = (3x+2)(x+2) has two different negative roots
k=8/3: a repeated root at x = -4/3
k=3: 3x^2+8x+6 has no real roots
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