Question

I AM SURE U WILL NOT BE ABLE TO SOLVE THIS EVEN IF U ARE POLYNOMIAL GENIUS........... CHALLENGE!???​

Answer 1

As we know,

B²-4AC=0

=(c-a)²−4(b-c)(a-b)=0

=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

=c²+a²−2ac-4ab+4ac+4b²-4bc=0

=c²+a²+2ac-4b(a+c)+4b²=0

=(a+c)²-4b(a+c)+4b²=0

=[(a+c)-2b]²=0

so, 2b=a+c.

Hence proved mark as brainliest

Answer 2

Roots are equal. So discriminant is zero.  

Therefore,  

$(c-a)^2-(4(b-c)(a-b)) = 0 \\ \\ (c^2-2ac+a^2)-(4(ab-b^2-ac+bc)) = 0 \\ \\ (c^2-2ac+a^2)-(4(ab+bc-ac-b^2)) = 0 \\ \\ (c^2-2ac+a^2)-(4ab+4bc-4ac-4b^2)=0 \\ \\ c^2-2ac+a^2-4ab-4bc+4ac+4b^2=0 \\ \\ 4b^2+a^2+c^2-4ab-4bc+2ac=0 \\ \\ (2b-a-c)^2=0 \\ \\ 2b-a-c=0 \\ \\ 2b=a+c$

Hence proved!!!

Thank you. Have a nice day. :-)  

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