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June 4th, 2019, 08:49 PM   #1
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Order of Operations with sqrt

Looking for step by step guide of how to answer. Have a “maybe” answer but it does not look right... struggling with +/- and square root.
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Last edited by skipjack; June 5th, 2019 at 08:12 AM.
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June 4th, 2019, 09:16 PM   #2
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$x = \pm b$ just means that $x$ can be either positive $b$ or negative $b$. Similarly, $x \pm b$ just means both $x+b$ and $x-b$.

Starting with:

$\displaystyle \frac{-20 \pm \sqrt{20^{2}-4(25)(4)}}{2(25)}$

$20^2 = 400$ and $4 \times 25 \times 4=400$

$\displaystyle \frac{-20 \pm \sqrt{400-400}}{2(25)}$

Now, the denominator, $2 \times 25 = 50$

$\displaystyle \frac{-20 \pm \sqrt{400-400}}{50}$

$400-400 = 0$ and $\sqrt{0}=0$

$\displaystyle \frac{-20 \pm 0}{50}$

Since its $\pm 0$ , adding and subtracting $0$ both do the same thing, so we will only have one $x$ value.

$\displaystyle \frac{-20 \pm 0}{50}=\frac{-20}{50} = \frac{-2}{5}$

So, we have $x=\frac{-2}{5}$

P.S: This appears to be the quadratic formula for $25x^{2}+20x+4$, so now that we know the $x$ value, we know $25x^{2}+20x+4 = 0$ at $x=\frac{-2}{5}$
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Last edited by Greens; June 4th, 2019 at 09:17 PM. Reason: LaTeX
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June 23rd, 2019, 06:44 AM   #3
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Quote:
Originally Posted by Greens View Post
$x = \pm b$ just means that $x$ can be either positive $b$ or negative $b$. Similarly, $x \pm b$ just means both $x+b$ and $x-b$.

Starting with:

$\displaystyle \frac{-20 \pm \sqrt{20^{2}-4(25)(4)}}{2(25)}$

$20^2 = 400$ and $4 \times 25 \times 4=400$

$\displaystyle \frac{-20 \pm \sqrt{400-400}}{2(25)}$

Now, the denominator, $2 \times 25 = 50$

$\displaystyle \frac{-20 \pm \sqrt{400-400}}{50}$

$400-400 = 0$ and $\sqrt{0}=0$

$\displaystyle \frac{-20 \pm 0}{50}$

Since its $\pm 0$ , adding and subtracting $0$ both do the same thing, so we will only have one $x$ value.

$\displaystyle \frac{-20 \pm 0}{50}=\frac{-20}{50} = \frac{-2}{5}$

So, we have $x=\frac{-2}{5}$

P.S: This appears to be the quadratic formula for $25x^{2}+20x+4$, so now that we know the $x$ value, we know $25x^{2}+20x+4 = 0$ at $x=\frac{-2}{5}$

I check this myself and I can confirm that it correct.-NinjaX3
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June 23rd, 2019, 01:11 PM   #4
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June 24th, 2019, 12:32 PM   #5
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I am glad that this helped.
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June 24th, 2019, 02:02 PM   #6
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Just to clarify (since the original title was order of operations), you can think of $\displaystyle \sqrt {20^2 - 4(25)(4)}$ as $\displaystyle (20^2 - 4(25)(4))^{0.5}$.

Taking the square root is just raising to a power, and takes the same priority. Simplify everything inside the radical, then take the root (if it simplifies), then continue on as normal (in this case, add or subtract from -20, and then divide by 50).
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