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The equation x4 - 2x3 - 3x2 + 4x - 1 = 0 has four distinct real rootsx1, x2, x3, x4 such that x1 < x2 < x3 < x4. Prove x1 x2 + x1 x3 + x2 x4 + x3 x4 = -3 |
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Answer» Given equation is x4 - 2x3 - 3x2 + 4x - 1 = 0 roots are x1, x2, x3, x4 such that x1 < x2 < x3 < x4. x1 x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4 = \(\frac{c}a = -3\)-----(i) Given that the product of two roots is units. Let x1x4 = 1 Also product of roots is x1 x2 x3 x4 = \(\frac{e}a=-1\) ⇒ x2 x3 = -1 (\(\because\) x2 x4 = 1 by assuming) \(\therefore\) From (i), we get x1 x2 + x1 x3 + 1 - 1 + x2 x4 + x3 x4 = -3 ⇒ x1 x2 + x1 x3 + x2 x4 + x3 x4 = -3 |
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