The equation \( x^{4}-2 x^{3}-3 x^{2}+4 x-1=0 \) has four distinct real roots \( x_{1}, x_{2}, x_{3}, x_{4} \)
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
6 views
1 Answers
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
6 views
Answered