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  • Brainly User
2015-04-18T16:42:13+05:30
We have the cubic equation x^{3}+13x^{2}+32x+20=0.
Clearly, it has three roots. Suppose that the roots are \alpha, \ \beta, \ \gamma.
Then,
\alpha+\beta+\gamma=-13 \\ \alpha \beta+\beta \gamma+\gamma \alpha=+32 \\ \alpha \beta \gamma=-20 .
Since one of the roots, say \gamma, is -2.
Then the above equations get modified into,
\alpha+\beta=-11 \\ \alpha \beta=10.
Then, | \alpha-\beta |= \sqrt{(\alpha+\beta)^{2}-4\alpha\beta}=\sqrt{(-11)^{2}-4(10)}=\sqrt{81}=9 .
Solving them you get the remaining roots to be \{-1, -10\} .
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