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Saturday, February 6, 2016

Quadratic equation: The relationship between solutions and coefficients 2

Question:
The equation, kx^2 -(k+3)x - 1 =0, which has real coefficients, has complex roots, a+ib and a-ib. In order to have those roots, find the range of k. If they are pure imaginary roots, find the value of k.

Answer:
The discriminant of the quadratic equation, ax^2+bx+c=0, is
D = b^2 - 4ac
For complex roots, we must have D<0 and k \neq 0. Thus,
\begin{eqnarray*} D &=& (k+3)^2 +4k    \\     &=& k^2+10k+9      \\     &=& (k+1)(k+9)  < 0 \end{eqnarray*}
The range of k is
-9 < k < -1 \quad \mathrm{but} \quad k \neq 0
If the roots are pure imaginary, we let the roots be ib and -ib. We have the following relationship:
ib + (-ib) = - \frac{-(k+3)}{k}
This is from the relationship between the roots and coefficients. Then,
\begin{eqnarray*} 0 &=& k +3  \\ \rightarrow k &=& -3 \end{eqnarray*}
This satisfies -9 < k < -1.

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