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Easy

Question

If alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 commathen the equation whose roots are alpha divided by beta straight & beta divided by alpha

  1. x to the power of 2 end exponent plus 5 x minus 3 equals 0    
  2. 3 x to the power of 2 end exponent plus 12 x plus 3 equals 0    
  3. 3 x to the power of 2 end exponent minus 19 x plus 3 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 equation whose roots are alpha divided by beta straight & beta divided by alpha.

The correct answer is: 3 x to the power of 2 end exponent minus 19 x plus 3 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 alpha not equal to beta comma alpha 2 equals 5 alpha minus 3 comma beta 2 times equals 5 beta minus 3 comma
    alpha squared equals 5 alpha minus 3
S i m p l i f y i n g space i t comma space w e space g e t colon
alpha squared minus 5 alpha plus 3 equals 0
beta squared equals 5 beta minus 3
S i m p l i f y i n g space i t comma space w e space g e t colon
beta squared minus 5 beta plus 3 equals 0
A s space alpha space a n d space beta space a r e space t h e space r o o t s comma space s o colon
x squared minus 5 x plus 3 equals 0
N o w space a s space p e r space t h e space c o n d i t i o n comma space w e space h a v e colon
alpha over beta plus beta over alpha
S i m p l i f y i n g space i t comma space w e space g e t colon
fraction numerator alpha squared plus beta squared over denominator alpha beta end fraction
fraction numerator left parenthesis alpha plus beta right parenthesis squared minus 2 alpha beta over denominator alpha beta end fraction
P u t t i n g space t h e space v a l u e s comma space w e space g e t colon
fraction numerator 25 minus 6 over denominator 3 end fraction
19 over 3
S o space t h e space e q u a t i o n space i s colon
x squared minus 19 over 3 x plus 1 equals 0
S i m p l i f y i n g space i t comma space w e space g e t colon
3 x squared minus 19 x plus 3 equals 0

    Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the equation is 3 x squared minus 19 x plus 3 equals 0.

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