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

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Joined: 22 Jul 2017
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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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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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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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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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