Maths-
General
Easy

Question

If α,β then the equationx squared minus 3 x plus 1 equals 0 with roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction will be

  1. x to the power of 2 end exponent minus x minus 1 equals 0    
  2. x to the power of 2 end exponent plus x minus 1 equals 0    
  3. x to the power of 2 end exponent plus x plus 2 equals 0    
  4. none of these    

hintHint:

The definition of a quadratic as a second-degree polynomial equation demands that at least one squared term must be included. It also goes by the name quadratic equations. Here we have to find the value of fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction if α,β are roots of the equationx squared minus 3 x plus 1 equals 0.

The correct answer is: x to the power of 2 end exponent minus x minus 1 equals 0


    A quadratic equation, or sometimes just quadratics, is a polynomial equation with a maximum degree of two. It takes the following form:
    ax² + bx + c = 0
    where a, b, and c are constant terms and x is the unknown variable.
    Now we have given the equation as: x squared minus 3 x plus 1 equals 0
    Here, a=1b=3 and c=1
    Now we have the sum of roots as α+β=-(-3)/1=3
    product of roots as α.β=1/1=1
    Adding the termsfraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction, we get :
    fraction numerator 1 over denominator alpha minus 2 end fraction plus fraction numerator 1 over denominator beta minus 2 end fraction
fraction numerator beta minus 2 plus alpha minus 2 over denominator left parenthesis alpha minus 2 right parenthesis left parenthesis beta minus 2 right parenthesis end fraction
fraction numerator beta plus alpha minus 4 over denominator alpha beta minus 2 alpha minus 2 beta plus 4 end fraction
N o w space u sin g space t h e space s u m space a n d space p r o d u c t s space w e space f o u n d space e a r l i e r comma space w e space g e t colon
fraction numerator 3 minus 4 over denominator 1 minus 2 left parenthesis 3 right parenthesis plus 4 end fraction
fraction numerator negative 1 over denominator 1 minus 6 plus 4 end fraction
fraction numerator negative 1 over denominator negative 1 end fraction equals 1 equals fraction numerator 1 over denominator alpha minus 2 end fraction plus fraction numerator 1 over denominator beta minus 2 end fraction

F o r space p r o d u c t comma space w e space g e t colon
fraction numerator 1 over denominator alpha minus 2 end fraction cross times fraction numerator 1 over denominator beta minus 2 end fraction
fraction numerator 1 over denominator left parenthesis alpha minus 2 right parenthesis left parenthesis beta minus 2 right parenthesis end fraction
fraction numerator 1 over denominator alpha beta minus 2 alpha minus 2 beta plus 4 end fraction
N o w space u sin g space t h e space s u m space a n d space p r o d u c t s space w e space f o u n d space e a r l i e r comma space w e space g e t colon
fraction numerator 1 over denominator 1 minus 2 left parenthesis 3 right parenthesis plus 4 end fraction
fraction numerator 1 over denominator negative 1 end fraction equals 1 equals fraction numerator 1 over denominator alpha minus 2 end fraction cross times fraction numerator 1 over denominator beta minus 2 end fraction

N o w space t h e space e q u a t i o n space w i l l space b e colon
x squared minus left parenthesis fraction numerator 1 over denominator alpha minus 2 end fraction plus fraction numerator 1 over denominator beta minus 2 end fraction right parenthesis x plus left parenthesis fraction numerator 1 over denominator alpha minus 2 end fraction cross times fraction numerator 1 over denominator beta minus 2 end fraction right parenthesis equals 0
P u t t i n g space t h e space v a l u e s space w h i c h space w e space f o u n d comma space w e space g e t colon
x squared minus left parenthesis 1 right parenthesis x plus left parenthesis negative 1 right parenthesis equals 0

    Here we used the concept of quadratic equations and solved the problem. We found the sum and product of the roots first and then proceeded for the final answer. Therefore, x to the power of 2 end exponent minus x minus 1 equals 0 will be the equation for the roots fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction .

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