what is the square root of -152?

√-152 = i√152 since √-1 = i

i√152 = i√4X38 = 2i√38





Rohn

The square root of any negative number is an imaginary number. It is some number multiplied by the "number" i, which i has the property that i^2 = -1. To solve your problem, take the square root of 152, then add 'i' to the end of it. There is no other way to deal with it, because of the fact that the square root of a negative number is imaginary.

Actually, complex numbers (part real, part imaginary) are useful, especially in solving differential equations and in analysis of sinusoidal waveforms of equal frequency. Even though the solutions are real, they have complex roots.

√-152 = √(-4*38)

= 2√-38

= 2√38*√-1

= (2√38)i , where i = √-1

Square Root Of 152

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RE:
what is the square root of -152?
i don't understand how to find the square root of a negative number. is it possible to explain without getting into real numbers and imaginary/complex numbers? they confuse me.

Hello tysongv, that is the beauty of mathematics. In our life we see real things around us but we imagine things. Hence the truth has got both real and imaginary. Thus the creation is complex.
Debate is going on about the existence of God and the spirits, ghosts etc. What is mind? That too is a riddle. One more interesting fact is that the whole creation is so measured and calculated. So mathematics is an unavoidable tool to unravel the secrets of Nature. Enjoy yourself and don't get scared and no need to get confused with complex numbers.
One example. What is cubeth root of 1? All of us would say without hesistation that the value is 1. But being third degree, it must have three roots. Some times we say all roots are equal and that too is 1. But complex world brings out the other two roots as complex conjugate roots.
They are (--1/2) +( /3 i/2) and (--1/2) -- (/3 i /2) [ /3 is to be treated as sqroot of 3]
Be cheerful. All the best.

√ (- 152 ) = √ ( 152 i ² ) = ± i √ (4 x 38 ) = ± 2 i √ 38

No need to be confused.
All you need to know is that i ² = - 1

√(-152)
= √(i^2 * 2^2 * 38)
= i * 2 * √38
= 2i√38

NB:
i = √(-1)

√-152 = (√-1)(√152)= i(√152) since i = √-1

i√152 = +12.329i or -12.329i

√(-152)

√(152)*i

4√(38)*i

2i*√(38)