Which Statement About Square Roots Of X-5 - Square X =5 Is True

What is the square root of 25?

Simple multiplication of 25*25 = 625.In Vedic mathematics, the numbers containing 5 in the units place can be squared by taking the remaining digit/s and multiply by its successor and then add 25.Square of 25 is 2*3 = 6 and attach 25, So 25^2 = 625.Square of 35 is 3*4 = 12 and attach 25, So 35^2 = 1225.Square of 125 is 12*13 = 156 and attach 25, So 125^2 = 15625.Square of 95 is 9*10 = 90 and attach 25, So 95^2 = 9025.From the time I had answered the question asking for the square of 25, the question now appears as what is the square root of 25.I will answer the revised question as square root of 25 is +5 or -5.

Is the square root of 9 = 3 or +-3?

Mathematics is really a language to communicate ideas. And like any other language, it only works if the words you use mean the same things to you that they mean to your audience. In fact, that idea is SO essential in mathematics that mathematicians are very careful to define every term we use that might be in any way ambiguous.Unfortunately, we sometimes get lazy. Sometimes, we use a shortened version of a term because we expect that the context of the statement will make our meaning clear to our audience.The most common definition of the term “square root” is as follows:A square root of a number [math]a[/math] is a number [math]y[/math] such that [math]y^2 = a[/math].Now, when we are talking about square roots of positive real numbers, [math]a[/math], we sometimes want to focus only the positive real number [math]y[/math] such that [math]y^2=a[/math]. This number is defined as the principal square root of [math]a[/math] and is denoted by the symbol [math]\sqrt a[/math].Now, as you might guess, mathematicians sometimes get lazy and refer to the principal square root as “the” square root. We expect our audience to know that when we say “the” we must mean a unique number, and what other number could we mean besides the principal square root? Shame on us for doing it, but we do it none-the-less.So despite our emphasis on communicating in an unambiguous way, you’ve hit on an example that is, indeed, ambiguous.Had you asked, “What are the square roots of of 9?” there would be no ambiguity. There are two square roots, [math]3[/math] and [math]-3[/math]. It is easy to show that both are square roots — all you need to do is square them both and observe that both yield [math]9[/math]. (It is not as easy to show that these are the only two square roots, but they are.)But you asked about “the square root of 9,” and we are left to guess whether you really meant to ask about “the square roots” or if you meant to ask about “the principal square root.” If you mean the second, then [math]3[/math] is the only answer. If you mean to ask about the first, then both the positive and negative answers are valid. If you don’t know which you meant, then hopefully you now understand the difference.

What is the square root of [math]28 x^2[/math] over [math]9[/math]?

Hello, Yes you are right, though this has confused alot of people, i will explain it as simple as it is:First, you should know:when you see [math]x[/math] (where here the largest degree is one) you conclude there will be only one root.[math]x^2[/math]  --> 2 Roots (there will be real or complex roots,double roots ... it doesn't matter ,after all there will be 2 roots)[math]x^3[/math][math] [/math]--> 3 Roots... etcNow the question that confuses a lot of people:Example : Solving... [math]x[/math][math]^2=4[/math]Leaves us with   [math]\sqrt{ (x^2)} = \sqrt{ (4)} = 2[/math]Which is same as your questionThe solution for this equation is obviously 2 and -2For [math]x=2[/math] --> [math]\sqrt{(x^2)}=\sqrt{(2^2)}= \sqrt{(4)}= 2[/math] {correct}For [math]x=-2[/math] --> [math]\sqrt{ (x^2) }= \sqrt {((-2)^2) }= \sqrt{ (4) }= 2 [/math]{correct}a simple explanation is that square roots never have negative value so they are absoluted  and[math]\sqrt{ (x^2) }= 2[/math]  is not same as  [math]x=2[/math]  {missed the second root}Actually [math]\sqrt{ (x^2)}=|x|[/math] wich can handle both positive and negative values So now we back on the right track-->[math]\sqrt{ (x^2) }= |x| [/math] and then [math]|x|[/math][math]=2[/math]Finally [math]x = 2[/math] or [math]x = -2[/math]In your question there's no equation so we will leave it ABSOLUTEDAnd the answer is  [math]\frac {2\sqrt{(7)}|x|}{3}[/math]And if you plan to put it in equation just restore back the absolute as [math]+-[/math]Now try this exapmle, let's see if you can find[math] x [/math]value (s) :1》 [math](x-2) ^2 = 1[/math]2》 [math](x+1) ^2 = 4[/math]Good-luck :)

What is x times the square root of x?

[math]x\sqrt{x}[/math][math]x^{1}x^{\frac{1}{2}}[/math][math]x^{\frac{2}{2}}x^{\frac{1}{2}}[/math][math]x^{\frac{3}{2}}[/math][math]\sqrt{x^{3}}[/math]In conclusion[math]x\sqrt{x}=\sqrt{x^{3}}[/math]

What is the square root of 49?

Intuitively, you are posing the following problem. A square has an area of 49. Of course, all the sides of the square have the same length. What is the length of the sides of the square?You can painstakingly arrange 49 small square tiles into a square shape to discover that you can make a [math]7 \times 7[/math] square with 49 tiles. Or, if you memorized the multiplication table at some point, you just know that [math]7 \times 7 = 49[/math].At a higher level, the square root function [math]f(x) = \sqrt{x}[/math] is defined on the domain [math][0, \infty)[/math], and is the inverse of the function of[math] g(x) = x^2[/math] over this domain. Wolfram Alpha, for instance, can tell you the value of [math]\sqrt{x}[/math] for (a great many values of) [math]x[/math] in this domain.Several people have answered that the square root of 49 is [math]\pm 7[/math]. This is incorrect; the square root function is a function, and therefore has a unique “output” for each “input.” However, it is true that the set of solutions to the equation [math]x^2 = 49[/math] is [math]x \in \{7,-7\}[/math]. This is perhaps a more subtle distinction than you might care to internalize, but it’s important if you’re actively doing math in some formal context.In general, up to multiplicity, there are [math]n[/math] solutions to an [math]n[/math]th degree polynomial as long as we allow for complex roots. This is a consequence of the Fundamental Theorem of Algebra, which states that a complex polynomial has a root. As an example, if we know that 7 is a root of [math]x^2–49=0[/math], then we can divide by [math]x-7[/math] to yield [math]x+7=0[/math], which has -7 as a root. Generally, if [math]a[/math] is a root of a polynomial equation [math]P(x) = 0[/math], we can divide [math]\frac{P(x)}{x-a}[/math] to get another polynomial with at least one root, so long as [math]P(x)[/math] has degree greater than 1. If [math]P(x)[/math] has degree one and [math]a[/math] is a root, then dividing [math]\frac{P(x)}{x-a}[/math] is effectively dividing [math]\frac{0}{0}[/math].

Is the square root of a number always smaller than that number?

The square root of a real number may be smaller than, equal to, or greater than the original number.If x is a member of the set of real numbers, the following statements are true:If x < 0, then √x will not be a real number. (√x will be a positive or negative unreal number, AKA imaginary number.) EG If x = –4, then √x= +√(–4) = +2 i AND-OR √x= —√(–4) = —2 i, such that i^2 = – 1If x = 0, then √x will be equal to x IE If x = 0, then √x= √0 = 0If 0 < x < +1, then one possible solution (+√x) will be greater than x ANDone possible solution (—√x) will be less than x EG If x = +1/9, then √x =+√(1/9) =+ 1/3. Note that 1/3 > 1/9 ……… AND-OR √x =—√(1/9) =— 1/3. Note that —1/3 < 1/9If x = +1, then √x will be equal to +/—x IE If x = 1, then √x = +√1 = +1 AND-OR √x = —√1 = —1 So one solution ( +1) equals x and the other solution (—1)is less than xIf x > +1, then √x =+/—√a where |x|=a Note that both +√a and —√a will be smaller than x EG If x = 16, then √x =+/— √16 = +/—4The graph of y = sqrt(x) is shown below, courtesy of http://www.desmos.comRegards,James2017 Dec 13 Wed