IV = V?

Algebra Level 2

I will attempt to prove that $4=5$ . Try and spot my mistake in the following steps:

A. Starting with the equality $16-36=25-45,$ B. we can add $\frac{81}{4}$ to both sides of this equality without changing anything: $16-36+\left( \frac { 81 }{ 4 } \right) =25-45+\left( \frac { 81 }{ 4 } \right).$ C. Using the perfect square trinomial properties, we can rewrite this equality as follows: ${ \left( 4-\frac { 9 }{ 2 } \right) }^{ 2 }={ \left( 5-\frac { 9 }{ 2 } \right) }^{ 2 }.$ D. Taking the square root of both sides, $4-\frac { 9 }{ 2 } =5-\frac { 9 }{ 2 }.$ E. Adding $\frac { 9 }{ 2 }$ to both sides gives $4=5.$

Which step contains the error?

No error in this proof C A D B E

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Victor Paes Plinio
Jan 10, 2015

The Answer is D because $\sqrt { { x }^{ 2 } } =\left| x \right|$ .

So:

$\sqrt { { \left( 4-\frac { 9 }{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-\frac { 9 }{ 2 } \right) }^{ 2 } }$

$\left| { \left( 4-\frac { 9 }{ 2 } \right) }^{ 2 } \right| =\left| { \left( 5-\frac { 9 }{ 2 } \right) }^{ 2 } \right|$

$\left| -0.5 \right| =\left| 0.5 \right|$

$0.5=0.5$

This is the correct way to solve this type of problem, the process in step D is wrong.

Correct!!!

PUSHPESH KUMAR - 6 years, 4 months ago

Yeah ur right.

chinmoy dutta - 6 years, 4 months ago

Chinmoy Dutta
Jan 15, 2015

√x^2 is never equal to x It is equal to |x|

IV = V?

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