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Find set of non real zero values of k such that | x^2-10x+9|=kx is satisfied by at least one and almost 3 values of x.

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Solution: We have , left | x^2-10x+9 
ight |=kx

               Rightarrow     x^2-10x+9=0Rightarrow x=9,1

              \Rightarrow x^2-10x+9=kx   Must have equal roots

              \Rightarrow x^2-x(10+k)+9=0,hspace1cmD=0\ \Rightarrow (10+k)^2-36=0Rightarrow k+10=pm 6\ \Rightarrow k=-16,-4

        And ,     \Rightarrow x^2-10x+9=-kx\ \Rightarrow x^2+x(k-10)+9=0

                      D=0Rightarrow k=16,4Rightarrow k=4  Almost  3 value of x

                 

Posted by

Deependra Verma

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