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In the equation a4x^4 + a3x^3 + a2x^2 + a1x + a0 =0 roots of equation are ai ,i =1,2,3,4. Now x is replaced by x – 1 ,now roots of new equation are

In the equation a4x^4 + a3x^3 + a2x^2 + a1x + a0 =0 roots of equation are ai ,i =1,2,3,4. Now x is replaced by x – 1 ,now roots of new equation are
 

Grade:11

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student
Please find the answer to your question below,
a_{4}x^{4} + a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} = 0
a1, a2, a3& a4are the root of this equation.
a_{4}(a_{i})^{4} + a_{3}(a_{i})^{3}+a_{2}(a_{i})^{2}+a_{1}(a_{i})+a_{0} = 0
Now for the equation,
a_{4}(x-1)^{4} + a_{3}(x-1)^{3}+a_{2}(x-1)^{2}+a_{1}(x-1)+a_{0} = 0
Put
x = a_{i} + 1
a_{4}(a_{i}+1-1)^{4} + a_{3}(a_{i}+1-1)^{3}+a_{2}(a_{i}+1-1)^{2}+a_{1}(a_{i}+1-1)+a_{0} = 0a_{4}(a_{i})^{4} + a_{3}(a_{i})^{3}+a_{2}(a_{i})^{2}+a_{1}(a_{i})+a_{0} = 0
So the roots of new equaion are
a1+1, a2+1,a3+1, a4+1

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