My Math Forum Order of Operations with sqrt

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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.

 June 4th, 2019, 09:16 PM #2 Member     Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry $x = \pm b$ 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 $x=\frac{-2}{5}$ Thanks from topsquark Last edited by Greens; June 4th, 2019 at 09:17 PM. Reason: LaTeX
June 23rd, 2019, 06:44 AM   #3
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Excellent

Quote:
 Originally Posted by Greens $x = \pm b$ 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

 June 23rd, 2019, 01:11 PM #4 Member     Joined: Oct 2018 From: USA Posts: 88 Thanks: 61 Math Focus: Algebraic Geometry Good news
 June 24th, 2019, 12:32 PM #5 Newbie   Joined: Jun 2019 From: New York Posts: 23 Thanks: 0 I am glad that this helped.
 June 24th, 2019, 02:02 PM #6 Senior Member   Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 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). Thanks from topsquark

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