Question

if α , β are the roots of the equations x²-2x+3=0 obtain the equation whose roots are α³-3α²+5α-2 , β³-β²+β+5

Answer 1

Given a , ß are the roots of the equation x^2 - 2x + 3 = 0

Given the roots of another equation are (a^3 - 3a^2 +5a-2) , (ß^3-ß^2+ß+5)

We have to find the new equation


x^2 -2x+3=0
x^2 +x-3x+3=0
x (x-1) - 3 (x-1) =0
(x-1)(x-3)=0
=> x-1=0 (OR) x-3=0
=> x=1 (OR) x=3

.°.The roots of the given equation are
1,3
.°. a=1 , ß=3

The roots of another equation are (a^3-3a^2+5a-2) , ( ß^3-ß^2+ß+5)
=>((1)^3-3 (1)^2 + 5 (1)-2) , ((3)^3-(3)^2+3+5)
=> (1-3+5-2) , (27-9+3+5)
=> 1 , 26
.°. 1,26 are the roots of new equation

General form of the quadratic equation is x^2-(a+ß)x+aß=0

For the new equation
Sum of the roots = 1+26 =27
Product of the roots = 1×26 =26

Quadratic equation is x^2-27x+26 =0

.°. The required quadratic equation is x^2 - 27x + 26 = 0



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