Case Positive!

Algebra Level 2

If x < 0 x<0 , then which of the following must be positive?

− x 2 -x^2 x 3 \sqrt[3]{x} − 2 x -2^{x} − x − 1 -x^{-1}

Anish Harsha by Anish Harsha , 20, India

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1 solution

Sravanth C.
Apr 8, 2016

When x < 0 x<0 it's value will be − ∣ x ∣ -|x| , substituting this value in each of the options we get:

1. − ∣ x ∣ 3 will be negative 2. − 2 − ∣ x ∣ = 1 − 2 ∣ x ∣ will be negative 3. − ( − ∣ x ∣ ) 2 = − ∣ x ∣ 2 will be negative 4. − ( − ∣ x ∣ − 1 ) = − 1 − ∣ x ∣ = 1 ∣ x ∣ will be positive \begin{aligned} 1. &\sqrt[3]{-|x|} &\text{will be negative} \\ 2. &-2^{-|x|} = \dfrac{1}{-2^{|x|}} &\text{will be negative} \\ 3. &-(-|x|)^2 = -|x|^2 &\text{will be negative} \\ 4. &-(-|x|^{-1}) = -\dfrac{1}{-|x|} = \dfrac 1{|x|} &\boxed{\text{will be positive}}\\ \end{aligned}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

In 3rd one − ( − x 2 ) = x 2 -(-x^2)=x^2 :ghost: :ghost: :ghost:

Nihar Mahajan - 5 years, 2 months ago

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hehe, edited. Thanks!

Sravanth C. - 5 years, 2 months ago

The first statement is wrong. If x<0 then -x>0. If you plug -x>0 to the first equation you're going to get a positive value. And last equation is going to be negative. Otherwise a good solution.

Henri Kärpijoki - 1 year, 2 months ago

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Ah, good catch. I've modified it accordingly :)

Sravanth C. - 1 year, 1 month ago

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