Square Roots are Tricky

Algebra Level 2

x = 4 \large x = \sqrt{4}

Find the value of x x .

2 - 2 ± 2 \pm 2 + 2 + 2

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2 solutions

Ram Mohith
Jun 22, 2018

Many of them think that the answer to this question ± 2 \pm 2 . But this is wrong answer. The correct answer is + 2 + 2 . I will explain you about this.

The actually definition of square root is below :

First, let us define a function y = x 2 y = \sqrt{x^2} . Now : { y = x if x < 0 y = x if x > 0 \begin{cases} y = \color{#20A900}x \text{ if } x < 0 \\ y = \color{#D61F06}- x \text{ if } x > 0 \\ \end{cases}

Now, we can write 4 \sqrt{4} as 2 2 \sqrt{2^2} and ( 2 ) 2 \sqrt{(-2)^2}

  • 2 2 = 2 (because 2 > 0) \sqrt{2^2} = \color{#20A900}2 \text{ (because 2 > 0)}

  • ( 2 ) 2 = ( 2 ) = 2 (because -2 < 0) \sqrt{(-2)^2} = \color{#D61F06}-(-2) = 2 \text{ (because -2 < 0)}

So, we have only one value for 4 \sqrt{4} that is + 2 + 2 .

Therefore, 4 = + 2 \color{#3D99F6} \text{Therefore, } \sqrt{4} = +2

I think the function should be f ( x ) f(x) instead of f ( y ) f(y) .

Henry U - 2 years, 3 months ago
Ritabrata Roy
Jul 23, 2018

Actually it's a definition. √a=|a|

Wrong! a 2 = a \sqrt{a^2}=|a|

Gia Hoàng Phạm - 2 years, 9 months ago

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