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nadya68 [22]
2 years ago
10

2x squared plus 16x minus 10 equals 0. please solve for the solution

Mathematics
2 answers:
ohaa [14]2 years ago
6 0
Answer: 5/9

Solution: 2x^2+16x-10=0

According to formula: X=(-16+_sqrt(16^2-(-10•2)) / 2•2

X=5/9
Semenov [28]2 years ago
4 0

Answer:

x = 0.58 or x = -8.58 to 2 d.p.

Step-by-step explanation:

Question:

Just to write the question in a better format :

2x^{2} +16x-10 = 0

Solve for x.

Answer:

First we Simplify this by dividing everything by 2:  

2x^{2} +16x-10 = 0 ÷ 2 :

x^{2} +8x-5 = 0

Now we use the quadratic formula :

x = \frac{-b+/-\sqrt{b^{2} -4ac} }{2a}  (side note: +/- is meant to be ±)

Now we substitute a with 1, b with +8 and c with -5:

x = \frac{-8+/-\sqrt{8^{2} -4(1)(-5)} }{2(1)}  x = \frac{-8+/-\sqrt{64+20} }{2} x = \frac{-8+/-\sqrt{84} }{2}  

The surd \sqrt{84} = 2\sqrt{21} So :

x = -8± 2√21 /2

x = -4 ±√21

This is our answer . Remember it is two answers in 1.

x = -4 +√21 and x = -4 -√21

x = 0.58257569495  or x = -8.58257569496

x = 0.58 or x = -8.58 to 2 d.p.

Hope this helped and brainliest please

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Step-by-step explanation:

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3 years ago
Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi
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Answer:

The general solution of   cos x = cos(\frac{\pi }{6})   is  

                                                x = 2nπ±\frac{\pi }{6}

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Step-by-step explanation:

Explanation:-

Given equation is  

                              2cosx-\sqrt{3} =0  for 0

                              2cosx =\sqrt{3}

Dividing '2' on both sides, we get

                             cos x =\frac{\sqrt{3} }{2}

                             cos x = cos(\frac{\pi }{6})

<em>General solution of cos θ = cos ∝ is θ = 2nπ±∝</em>

<em>Now The general solution of   </em>cos x = cos(\frac{\pi }{6})<em>   is  </em>

<em>                                                 x = 2nπ±</em>\frac{\pi }{6}<em></em>

put n=0

x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Put n=1  

x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}

x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}

put n=2

x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}

x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}

And so on

But given 0 < x< 2π

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

                               

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