Problem Solving (PS) | Re: If x is a prime number and |a + b + c - 2x| + (x^2 - a - b - c)^2 =

 

\( |a + b + c - 2x| + \sqrt{(x^2 - a - b - c)^2} = -|c|\)

Rule: \( \sqrt{(m)^2} =|m|\)

\( |a + b + c - 2x| + |(x^2 - a - b - c)| = -|c|\)

The left hand side of the equation must be a non-negative number as we are adding two values which are non-negative.

Hence, the value of\(c\) cannot be positive and must bezero.

\( c =0\)

\( |a + b + c - 2x| + |(x^2 - a - b - c)| =0\)

As the sum equals zero, each of the term must be equal tozero

\( a + b + c - 2x =0\) and \( x^2 - a - b - c =0\)

\( a + b + c - 2x =0\)

\( a + b + c = 2x\) →(1)

\( x^2 - a - b - c =0\)

\( x^2 = a + b + c\)  → (2)

(1) =(2)

\( x^2 =2x\)

As\(x\) is a prime number, we know that\( x \neq0\) , hence dividing both
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Statistics : Posted by PrabhatKC • on 14 May 2024, 01:30 • Replies 3 • Views 190

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