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dexar [7]
2 years ago
13

Solve for x: 3√5x² +25x-10√√5 = 0

Mathematics
2 answers:
bogdanovich [222]2 years ago
3 0

Answer:  x = - 2√5, √5/3

Step-by-step explantion

3 √ 5 x 2 + 25 x − 10 √ 5 = 0 35x2+25x−105=0

⇒ 3√5x2 + 30x – 5x – 10√5 = 0  

⇒ 3√5x(x + 2√5) – 5(x + 2√5) = 0  

⇒ (x + 2√5)(3√5x – 5) = 0

x = - 2√5, √5/3

Natasha2012 [34]2 years ago
3 0

3 \sqrt{5}  {x}^{2} +  25x - 10 \sqrt{5}  = 0 \\  \\ 3 \sqrt{5} {x}^{2}   + 30x - 5x - 10 \sqrt{5}  = 0 \\  \\ 3x( \sqrt{5} x + 10) -  \sqrt{5} ( \sqrt{5}  + 10) = 0 \\  \\ ( \sqrt{5} x + 10)(3x -  \sqrt{5} ) = 0

\sqrt{5} x  + 10 = 0 \\  \\  \sqrt{5}x =  - 10 \\  \\ x =   \frac{ - 10}{ \sqrt{5} }   \\  \\ x =  \frac{ - 2 \times \sqrt{5 }  \times  \sqrt{5} }{ \sqrt{5} }  \\  \\ x = -  2 \sqrt{5} .

3x -  \sqrt{5}  = 0 \\  \\ 3x =  \sqrt{5}  \\  \\ x =  \frac{ \sqrt{5} }{3} .

The values of x are -2√5 & √5/3 .

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Answer:

1. x = +/- 2\sqrt{2} - 7

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Square root both sides: x + 7 = +/- 2\sqrt{2}

Subtract 7 from both sides: x = +/- 2\sqrt{2} - 7

2. Square root both sides: x - 3 = \sqrt{-12}

Since there is a negative inside the radical, we need to have an imaginary number: i=\sqrt{-1} . So, \sqrt{-12} =i\sqrt{12} =2i\sqrt{3}

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3. Divide by -5 from both sides: (n - 2)^2 = -2

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Hope this helps!

8 0
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