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# If M=√2÷√(×+3)(×-2) determine the value of × for which M is real

if M=√2×÷√(×+3)(×-2) determine the value of x which M is real

## Research, Knowledge and Information :

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### if the equation below is to have 1 double real root ...

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### #4x^2-6mx+(2m+3)(m-3)# - Socratic.org

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### Linear Functions Part A: Basics: Slope and Intercept

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### mid ans dvi - Northwestern University

(2) Determine the truth value of each of these statements if the universe ... (3) (a) TRUE (√ 2 is a real number.) (b) ... mid_ans_dvi Created Date: Wed May ...

### Chem Final All of the Problems Flashcards | Quizlet

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Section 2.1: BASIC PROOFS WITH QUANTIFIERS Existence ... by the Intermediate Value Theorem, there is a real number r ... Example 2: Prove that for every real number x ...

## Suggested Questions And Answer :

### minimun of sqrt(x^4-3x^2+4)+sqrt(x^4-3x^2-8x+20)

f = sqrt(x^4-3x^2+4)+sqrt(x^4-3x^2-8x+20) To find the turning points, differentiate f and set that value to zero. df/dx = (1/2)*(4*x^3-6*x)/sqrt(x^4-3*x^2+4)+(1/2)*(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20) = 0 (4*x^3-6*x)/sqrt(x^4-3*x^2+4) = -(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20) cross-multiplying and squaring, (4*x^3-6*x)^2*(x^4-3*x^2-8*x+20) = (4*x^3-6*x-8)^2*(x^4-3*x^2+4) expanding this gives, 16*x^10-96*x^8-128*x^7+500*x^6+384*x^5-1068*x^4-288*x^3+720*x^2 = 16*x^10-96*x^8+244*x^6-236*x^4-64*x^7+288*x^5-544*x^3-48*x^2+384*x+256 which simplifies as, 64*x^7-256*x^6-96*x^5 +832*x^4-256*x^3-768*x^2+384*x+256 = 0 2x^7 - 8x^6 - 3x^5 + 26x^4 - 8x^3 - 24x^2 + 12x + 8 = 0 which factorises as, (x^2 - 2)(2x^5 - 8x^4 + x^3 + 10x^2 - 6x - 4) = 0 Now the (x^2 - 2) term gives you the two roots, +/- sqrt(2) The quintic equation has 3 real roots and two complex roots. The real roots are at (approx) x = -1, x = -0.5, x = 3.5. (I know where the roots are because I graphed the quintic and read from the graph.) If you wish to get these roots more accurately, then use the Newton-Raphson method with the above root values as strarting values. The fastest way of finding the minimum of the function, f, is to plug in all the above root values and determine the minimum value from that. A faster method would be to graph the curve of f(x), then say, "From the graph of f(x) it can be seen that the smallest local minimum is at where x lies between x = 1 and x = 2. From the solution, x^2 - 2 = 0, we can say that the mimimum occurs at x = sqrt(2)." Then include the minimum value. f(sqrt(2)) = sqrt(4 - 6 + 4) + sqrt(4 - 6 - 8sqrt(2) + 20) sqrt(2) + sqrt(18 - 8sqrt(2)) sqrt(2) + sqrt(16 - 8sqrt(2) + 2) sqrt(2) + sqrt{(4 - sqrt(2))^2} sqrt(2) + 4 - sqrt(2) f(sqrt(2)) = 4 I'm afraid that I couldn't find all the real roots analytically, so this solution is a bit incomplete, however you may still find it useful.

### Basic functions

1. What is the range of k(x)=-|x| Asking for the range is another way of asking for valid Y values (or in this case k(x) values). So the range is (-infinity, 0]. 2.Is y=2 a constant function? Yes. Constant functions have no variation in the output for any input. 3. x=-3 a function? No it is not. In order to be a function, there are several properties which must be satisfied, one of which is that for any input, there must be a single output. Often we simplify this by asking "does the function pass a vertical line test?" So if you are to draw a vertical line, would it only cross the function in one place for all values of X? For X=-3, it is a vertical line, so it fails this property of functions because there are an infinite number of solutions (Y values) for this single X value (-3). 4. What is the minimum value of y=x^2-4 The minimum value of parabola's will occur either at the vertex (where its slope is 0) or at its domain boundaries. There are no stated domain restrictions for this function and the slope is 0 at X=0. The first derivative is y'=2x and y'=0 at x=0. The second derivative is y''=2, indicating it is concave for all values X. So the minimum is x=0 and y=-4. For a function with a higher power, there may be more than one minimum (each called local minimum) and any place the second derivative is concave up, both to the left and right segments where the first derivative is 0 will be one of these local minimums. You will take your inputs for each relative minimum and any domain limits (for functions which do not have all real numbers for their domain) and compare the Y values of the original function to determine the absolute minimum for that function.

### If M=√2÷√(×+3)(×-2) determine the value of × for which M is real

This is easier than it looks. The square root term including x is only meaningful in this context when the quantity under the square root is strictly positive, so (x+3)(x-2)>0.  So we have just 2 conditions when this is positive: x+3 and x-2 are both positive; and x+3 and x-2 are both negative. Both positive means x+3>0 and x-2>0, so x>-3 and x>2. Since 2>-3, x>2 satisfies both inequalities.  Now when are they both negative? x+3<0 and x-2<0, so x<-3 and x<2. Since -3<2 then x<-3 satisfies both inequalities. So there you have it: M is real when x>2 or x<-3. But if √2x is the numerator rather than √2 and the expression is √(2x) rather than (√2)x, we must eliminate x<-3 leaving x>2 as the solution.

### find the real root of x^3-5x+1 by regula-falsi method

find the real root of x^3-5x+1 by regula-falsi method. f(x) = x^2 - 5x + 1 Test for a root f(0) = 1, f(1) = -3. There is a change of sign between f(0) and f(1) , so there is a root between x = 0 and x = 1. The first (smallest) false root is denoted as a, the second (greatest) false root is denoted by b, the interpolated (false) root is denoted by c and c is given by, c = (a.f(b) - b.f(a)) / (f(b) - f(a)) After c is determined, it will replace either a or b for the next iteration. The sign of f(c)  is found, and c will replace a if f(c) has same sign as f(a), otherwise it will replace b. a b f(a) f(b) c=(a.f(b) - b.f(a)) / (f(b) - f(a)) sign(f(c)) Regula-falsi method 0 1 1 -3 0.25 -ve 0 0.25 1 -.234375 0.2025668264 -ve 0 0.2025668264 1 -0.4522143e-2 0.2016336225 +ve 0.2016336225 0.2025668264 0.0000295283 -0.4522143e-2 0.2016396263 +ve 0.2016396263   2.406*10^(-7)       When c reaches the value 0.2016396263, f(c) = 2.406*10^(-7), which is small enough to allow c to be taken as a real root of f(x). Root of f(x) is: c = 0.2016396263

### solve for y=3x+4, and show both ordered sets on graph

The equation is linear in two variables, x and y. It can't be solved, as such, for particular values of x and y, but a straight line graph represents the dependency of one variable on the other. The graph is a mapping of the function in which the range of values of x and y is the set of all real numbers, so the graph maps x values onto y and y onto x in a unique way: there is only one value of x for a particular value of y, and only one value of y for a particular value of x. There is a real value of x for every real value of y and vice versa, subject to the dependency function y=3x+4, and no real values are excluded. The ordered pair (x,y) forms the set of real numbers satisfying the equation. Every point on the line (and there are an infinite number of them) is an ordered pair, and forms an infinite set. Each ordered pair is effectively a solution to the equation, so there are an infinite number of solutions.

### Mathmatical sequences

i have a sequence that is: 60,57,51,42 what are the next three numbers

### If 2 + 22 = 15 & 4 + 16 = 16 & 6 + 10 = 17 & 8 + 4 = 18 what does 10 + 10 =

2 + 22 = 15, 4 + 16 = 16, 6 + 10 = 17 and 8 + 4 = 18   There must be certain common multiples, x and y, that make the real value of left-hand side(LHS) in each expression above equal to the real value of its right-hand side(RHS): x for the 1st terms of each LHS, { 2, 4, 6, 8 }, and y for the 2nd terms of each LHS, { 22, 16, 10, 4 }. So that, the expressions above can be rewritten in their real values, assuming the number of each RHS shows the real value of its LHS: 2x + 22y = 15 ··· Eq.1,  4x + 16y = 16 ··· Eq.2,  6x + 10y = 17 ··· Eq.3, 8x + 4y = 18 ··· Eq.4.    From Eq.1 and Eq.2, we have: x = 2 and y = 0.5. CK: Plug these values of x and y into each LHS of Eq.1, Eq.2, Eq.3 and Eq.4.   2·(2) + 22·(0.5) = 15,    4·(2) + 16·(0.5) = 16,   6·(2)+10·(0.5) = 17 and 8·(2) + 4·(0.5) = 18   CKD. Thus, in each given expression, the 1st term of LHS indicates 2 times of its number, the 2nd term of LHS indicates one half of its number, and the number of RHS expresses the real value of LHS.  Therefore, the answer is 10 + 10 = 25  {=10·(2)+10·(0.5)}