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How to solve: 5x^2-34x+24=0

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How to solve: 5x^2-34x+24=0  [#permalink]

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New post 16 Dec 2018, 23:09
Hi, can anyone walk me through the steps to solve:

5x\sqrt{2}-34x+24=0

Thanks!
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Re: How to solve: 5x^2-34x+24=0  [#permalink]

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New post 16 Dec 2018, 23:50
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Let's compare the given equation to the standard equation ax^2+bx+c

This means,a=5,b=34,c=24.

Now,find 2 factors of a*c which sum up to b.

a*c= 5*24=120.

Factors are 30 and 4
30*4=120

And 30+4=34

Thus,the new equation can be written as 5x^2-30x-4x+24=0
Or (x-6)(5x-4)=0

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Re: How to solve: 5x^2-34x+24=0  [#permalink]

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New post 18 Dec 2018, 08:43
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gempony wrote:
Hi, can anyone walk me through the steps to solve:

5x\sqrt{2}-34x+24=0

Thanks!


\(5x^2 - 34x + 24=0\)

Or, \(5x^2 - x ( 30 + 4 )+ 24=0\)

Or, \(5x^2 - 30x -4x + 24=0\)

Or, \(5x (x - 6) - 4( x - 6)=0\)

So, Either \(5x - 4 = 0\), ie \(x = \frac{4}{5}\) OR, \(x - 6 = 0\) , So \(x = 6\)
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Re: How to solve: 5x^2-34x+24=0  [#permalink]

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New post 21 Dec 2018, 16:19
gempony wrote:
Hi, can anyone walk me through the steps to solve:

5x\sqrt{2}-34x+24=0

Thanks!


In addition to what's already been said above, you can use the quadratic formula. It gives you two values for x:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In this case, it would take a bit of arithmetic to use it, but here's how it works out:

\(x = \frac{34 \pm \sqrt{34^2 - (4)(5)(24)}}{10}\)

\(x = \frac{34 \pm \sqrt{1156 - 480}}{10}\)

\(x = \frac{34 \pm \sqrt{676}}{10}\)

\(x = \frac{34 \pm 26}{10}\)

\(x = \frac{60}{10} = 6\) or \(x = \frac{8}{10} = \frac{4}{5}\)
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Re: How to solve: 5x^2-34x+24=0   [#permalink] 21 Dec 2018, 16:19
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