Question 173793
If I say the roots are {{{r[1]}}} and {{{r[2]}}}, then
{{{(x - r[1])(x - r[2]) = x^2 -(r[1] + r[2])x + r[1]r[2]}}}
given the equation
{{{kx^2 + (k-1)x + 2k + 3 = 0}}}
divide both sides by {{{k}}}
{{{x^2 + ((k-1)/k)*x + (2k + 3)/k = 0}}}
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(a) one root is the negative of the other.
{{{r[1] = -r[2]}}}
{{{r[1] + r[2] = 0}}}
therefore the coefficient of {{{x}}} is zero
{{{(k - 1)/k = 0}}}
{{{k = 1}}}
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{{{r[1] = 1/r[2]}}}
{{{r[1]*r[2] = 1}}}
{{{(2k+ 3)/k = 1}}}
{{{2k + 3 = k}}}
{{{k = -3}}}
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(c) one root is the twice of the other.
{{{r[1] = 2*r[2]}}}
{{{-(r[1] + r[2]) = -3*r[2]}}}
{{{-3*r[2] = (k - 1)/k}}}
{{{r[2] = (1 - k)/(3k)}}}
{{{r[1]*r[2] = 2*(r[2])^2}}}
{{{2*(r[2])^2 = (2k + 3)/k}}}
{{{2*((1 - k)/(3k))^2 = (2k + 3)/k}}}
{{{(4*(1-k)^2)/ (9k^2) = (2k +3)/k}}}
{{{4k*(1-k)^2 = 9k^2*(2k + 3)}}}
{{{4k*(1 - 2k + k^2) = 18k^3 + 27k^2}}}
{{{4k - 2k^2 + 4k^3 = 18k^3 + 27k^2}}}
{{{14k^3 + 29k^2 - 4k = 0}}}
{{{k*(14k^2 + 29k - 4) = 0}}}
{{{k = 0}}} and the solution to the quadratic
(unless I goofed)