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  • Solve! - Complex numbers and quadratic equations - JEE Main-8

The sum of all real values of x satisfying the equation

\left ( x^{2} \right-5x+5 )^{x^{2}+4x-60}=1

is

 

  • Option 1)

    3

  • Option 2)

    -4

  • Option 3)

    6

  • Option 4)

    5

 

Answers (1)

As we have learned

Roots of Quadratic Equation with real Coefficients -

\alpha ,\beta are roots if

ax^{2}+bx+c= 0

is satisfied by x= \alpha ,\beta

 

- wherein

\alpha ,\beta\in C

a,b,c\in R

 

 

 (x^2 -5x +5 )^{x^2+4x-60}= 1 \\ \Rightarrow x^2 -5x +5 = 1 \: \: or \: \: x^2+4x-60=0

\Rightarrow x^2 -5x +4 = 0 \: \: or \: \: x^2+4x-60=0

\Rightarrow (x-4)(x-1)= 0 \: \: or \: \: (x+10)(x-6)= 0

x= 4, 1 \: \: or \: \: x = 6, -10

Also there is a case 

x^2 -5x +5 = -1 \: \: \: \& \: \: \: x^2+4x-60=even \: \: no.

x^2 -5x +6 = 0

(x-3)(x-2)= 0 

= x = 2,3 

For x = 3 : x^2 +4x -60 = -39 (not \, \, \, accepted )

So sum 4+1+6+(-10)+2= 3 

 

 

 

 

 

 


Option 1)

3

Option 2)

-4

Option 3)

6

Option 4)

5

Posted by

Vakul

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