Solving for the parameter.

Algebra Level 3

Given the equation x 2 − 2 ( m + 1 ) x + m 2 = 0 x^2-2(m+1)x+m^2=0 , where m m is the parameter.

Find all real values of m m such that the equation above has 2 real solutions x 1 , x 2 x_1,x_2 satisfying ( x 1 − m ) 2 + x 2 = m + 2 (x_1-m)^2+x_2=m+2 .

Type your answer as the sum of all solutions. If there is no solution, type 0 0 .


The answer is -0.5.

Tin Le by Tin Le , 15, Vietnam

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1 solution

There are two possibilities :

(i) 2 m + 1 ( 1 + 2 m + 1 ) = 0 \sqrt {2m+1}(1+\sqrt {2m+1})=0 , which has one solution :

m = − 1 2 m=-\dfrac 12

(ii) 2 m + 1 ( 2 m + 1 − 1 ) = 0 \sqrt {2m+1}(\sqrt {2m+1}-1)=0 , which has two solutions :

m = 0 , − 1 2 m=0,-\dfrac 12

So, sum of all distinct values of m m is − 1 2 = − 0.5 -\dfrac 12=\boxed {-0.5} .

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