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If z is negative, then [m]\sqrt[4]{(4z-5)^4} + [square_root](3-2z)^2[/

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If z is negative, then [m]\sqrt[4]{(4z-5)^4} + [square_root](3-2z)^2[/ [#permalink] New post  17 Feb 2023, 23:07
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If z is negative, then \(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z |z|} = ?\)

A. 3z-2
B. z-2
C. -8
D. 8-7z
E. 4z-8

Source: Top One Percent (Sandeep Gupta)
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If z is negative, then [m]\sqrt[4]{(4z-5)^4} + [square_root](3-2z)^2[/ [#permalink] New post  18 Feb 2023, 01:28
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↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧




Given that z is negative and we need to find the value of \(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z |z|}\)

Let z = -k (where k is positive) (k = -z)

=> \(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z |z|}\) = \(\sqrt[4]{(4(-k)-5)^4} + \sqrt{(3-2(-k))^2} + \sqrt{-(-k) |(-k)|}\) = \(\sqrt[4]{(-(4k+5))^4} + \sqrt{(3+2k)^2} + \sqrt{k*k}\)
( As modulus of |-k| = k for k > 0 and modulus of any number is the non-negative value of that number)

Watch this video to know the Basics of Absolute value

=> \((4k+5)^{\frac{4}{4}} + 3+2k + k\)
=> 4k+5 + 3 + 3k = 7k + 8 = 7*(-z) + 8 = 8 - 7z

So, Answer will be D
Hope it helps!

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If z is negative, then [m]\sqrt[4]{(4z-5)^4} + [square_root](3-2z)^2[/ [#permalink] New post  18 Feb 2023, 07:40
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Kushchokhani wrote:
If z is negative, then \(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z |z|} = ?\)

A. 3z-2
B. z-2
C. -8
D. 8-7z
E. 4z-8

Source: Top One Percent (Sandeep Gupta)


Concept Tested :
  • \(\sqrt{x^2} = |x|\)
  • |x| = x , when \(x \geq 0\)
  • |x| = -x , when \(x < 0\)

\(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z |z|}\)

As z is negative, |z| = -z

\(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{-z (-z)}\)

\(\sqrt[4]{(4z-5)^4} + \sqrt{(3-2z)^2} + \sqrt{z^2}\)

\(|4z-5| + |3-2z| + |z|\)

We know that z < 0

⇒ 4z - 5 < 0, hence |4z-5| = -(4z-5) = 5 - 4z

⇒ 3-2z > 0, hence |3-2z| = 3-2z

⇒ |z| = -z

Substitute the values

\(|4z-5| + |3-2z| + |z|\) = \(5 - 4z + 3 - 2z - z\)

= \(8 - 7z\)

Option D
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If z is negative, then [m]\sqrt[4]{(4z-5)^4} + [square_root](3-2z)^2[/ [#permalink]
18 Feb 2023, 07:40
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