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NISA [10]
4 years ago
13

Simplify square root of negative 48

Mathematics
1 answer:
77julia77 [94]4 years ago
5 0
\bf \sqrt{-48}\qquad 
\begin{cases}
48=2\cdot 2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 2^2\cdot 3\\
\qquad (2^2)^2\cdot 3
\end{cases}\implies \sqrt{-1\cdot (2^2)^2\cdot 3}
\\\\\\
\sqrt{-1}\cdot \sqrt{(2^2)^2\cdot 3}\implies i\cdot 2^2\sqrt{3}\implies 4\sqrt{3}\ i
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Answer:

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Step-by-step explanation:

Let the two numbers be 'x' and 'y'.

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The squares of the sum of two numbers = 144

The sum of their squares = 180

The sum of the numbers = x+y

The square of the sum of numbers = (x+y)^2

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The sum of their squares = x^2+y^2

Now, as per question:

(x+y)^2=144---(1)\\\\x^2+y^2=180---(2)

Expanding equation (1) using the formula (a+b)^2=a^2+b^2+2ab

This gives,

(x+y)^2=144\\\\x^2+y^2+2xy=144\\\\\textrm{Plug in the value of }x^2+y^2\textrm{ from equation (2)}\\\\180+2xy=144\\\\2xy=144-180\\\\2xy=-36\\\\xy=-\frac{36}{2}\\\\xy=-18\\\\y=-\frac{18}{x}---(3)

Plug in the value of 'y' from equation (3) in equation (1). This gives,

x^2+(-\frac{18}{x})^2=180\\\\x^2+\frac{324}{x^2}=180\\\\x^4+324=180x^2\\\\x^4-180x^2+324=0\\\\\textrm{On solving, we get:}\\\\x^2=178.2\ or\ x^2=1.8\\\\\textrm{Square root both sides, we get:}\\\\x=\pm13.35\ or\ x=\pm1.35

Therefore,

y=-\frac{18}{x}=-\frac{18}{\pm13.35} = \pm1.35\ or\\\\y=-\frac{18}{x}=-\frac{18}{\pm1.35}=\pm13.3

Therefore, the numbers are either ±13.35 and ±1.35 or ±1.35 and ±13.3

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