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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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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 Tags operations, order, sqrt Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post porknbeans Elementary Math 5 November 14th, 2014 06:33 PM fran1942 Algebra 3 May 8th, 2012 12:55 AM melorock089 Calculus 2 September 22nd, 2009 04:36 AM floaty Elementary Math 3 January 7th, 2009 07:43 AM Lovelace Algebra 2 December 13th, 2007 01:56 PM

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