Guide :

how do i solve -1 = - (square root of x)

original equation f(x) = 3 - square root of x   and f(x) = 2

Research, Knowledge and Information :


How to Solve Square Root Problems (with Pictures) - wikiHow


How to Solve Square Root Problems. ... square root problems are not as hard to solve as they may first seem. Simple square root problems can ...
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Solve Equations With Square Root - analyzemath.com


Solve Equations With Square Root. Tutorial on how to solve equations containing square roots. Detailed solutions to examples, ... x + 1 = 16 Solve for x. x = 15
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Solving square-root equations (basic) (video) | Khan Academy


Solving square-root equations: one solution. ... We're asked to solve for x. So we have the square root of the entire quantity 5x squared minus 8 is equal to 2x.
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Squares and Square Roots in Algebra - Maths Resources


A square root of x is a number whose square is x: ... Two Square Roots. And that means ..... a square root of 25 can be 5 or −5. ... Solve x 2 − 9 = 0 : Start with:
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Solving Radical Equations: Examples - Purplemath


Demonstrates how to solve equations containing multiple square-root ... Solve: Since there is a square root inside ... "Solving Radical Equations: Examples."
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How do I solve this math problem?_ the square root of 3x + 1 ...


How do I solve this math problem? the square root of 3x + 1 = 3 + the square root of x - 4. Answers ... x = 9 - 5. Step 4- Solve the equation. 2x = 4. Step 5. x = 4/2.
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9.4 using square root q property to solve ... - WOU Homepage


PROPERTY TO SOLVE QUADRATIC EQUATIONSQUADRATIC EQUATIONS. 6/page to print. ... square’, and apply the squareand apply the square root property to solve quad.eq. x. 2
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Solving Radical Equations LINK - Math is Fun


Solving Radical Equations. How to solve equations with square roots, ... Solving Radical Equations. ... (x−1) Now do the "square root" thing again:
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Suggested Questions And Answer :


how to solve this quadratic equation x^2-2^x-13=0

I think you may have typed this in wrongly because the term 2^x is not a permitted term in a quadratic equation. As it stands the solution to the equation is about -3.6168. Let's say you made a mistake in typing and the middle term is 2*x or 2x, then the solution exists but it is irrational (involves square roots). The easiest way to solve x^2-2x-13=0 is to complete the square: (x^2-2x+1)-1-13=0, which is (x-1)^2-14=0, so (x-1)^2=14. Take square roots of each side: x-1=+sqrt(14)=+3.7417 approx. A square root always has two solutions, one positive and one negative, but the same magnitude number. So x has two possible values: x=1+3.7417=4.7417 or x=1-3.7417=-2.7417. If you meant the question to be x^2-12x-13=0, then the solution is (x-13)(x+1)=0 and x=13 or -1. To work this out we ask: what are the factors of 1 (x^2 coefficient) and 13 (the constant term). The factors of 1 are 1 and 1 because only 1 times 1 make 1; and the factors of 13 are only 1 and 13. The coefficient of the x^2 term tells is how many x's go in each bracket. That's 1x or just x in each bracket. And the factors of 13 tell us what. Numbers to write in each bracket, so that's 1 and 13. So we have (x 1)(x 13). What about the signs between them? We look at the sign in front of 13 in the quadratic. It's minus, and that means there will be a plus in one bracket and a minus in the other. But which way round? Well, there's one more test: we take the factors of 13 and subtract them because the minus sign in front of 13 tells us we need to subtract. If it had been plus, we would have added the factors. 13-1=12. If 12 is the coefficient of the x term then the quadratic can be solved. (If the number had not been 12 we could not have solved the quadratic this way.) The sign in front of 12x is the sign that goes in front of the larger number in the brackets, so minus goes in front of 13. So we have (x+1)(x-13)=0. One or other of these factors is zero, so x+1 or x-13 is zero. x+1=0 means x=-1 and x-13=0 means x=13. These are the solutions or roots.
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explain how to solve: square root of 5x - 3 - square root 10 - 4x =0

Sqrt(5x-3)-sqrt(10-4x); sqrt(5x-3)=sqrt(10-4x); Square both sides: 5x-3=10-4x; 9x=13 so x=13/9.
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How would I solve the following equation by completing the Square? x^2-18x+86=0

How would I solve the following equation by completing the Square? x^2-18x+86=0 I just need to know how to solve it, it's Algebra 2 math homework and its on completing the square! When completing the square, you move the constant to the right side of the equation; in this case you do that by subracting it from both sides. That leaves the x squared and x terms on the left side of the equation. Divide the co-efficient of the x term by the co-efficient of the x squared term. It's easy when the co-efficient of the x squared term is 1. You simply end up with the co-efficient of the x term. Take half of that and square it. Add that number to both sides of the equation. You now have to take the square root of both sides of the equation. On the left side, you will have  x plus one-half of the x term. On the right side perform the addition and take the square root of the result. In general terms, you will now have something like (x + d) = sqrt (h). d may be negative if the co-efficient of the x term was negative. Add or subtract d to/from both sides of the equation, leaving x all by itself of the left side.
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square root7under square root7 under square root 7

Let y=√(7√(7√7)), then y^2=7√(7√7); y^4=7^2*7√7=7^3√7; y^8=7^6*7=7^7. So y=7^(7/8)=5.4886 approx. In general, if N is the number of 7's under the square root then y=7^(((2^N)-1)/2^N). In this case N=3 so y=7^((2^3 - 1)/2^3)=7^(7/8). For N=4, y=7^(15/16)=6.1984; as N approaches infinity y approaches 7. So for 7's repeated under square roots indefinitely the answer is 7. Another way of solving it is to put y=√(7(√7(√7... So y^2=7√7(√7√7... =7y, so since y≠0, y=7 (dividing through by y).
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x^2-18x=-56

Alright remember every quadratic equation can be put in the for ax^2 +bx +c = 0 will I be referring to a,b,and c in this explanation. To solve by completing the square we to move the c to the other side which your problem already has done. We then need to take your b term whick is -18 and divide it by two and square it. We will do that step every time when completing the square so just remember it (b/2)^2. We get 81. We add 81 to both sides. Now we have x^2 -18x +81 = -51 +81 we can simplify this to (x-9)^2 = -30 Most students get confused on how I obtained (x - 9) ^2. The best way I can explain it is when completing the square, out objective is to get a perfect square binomial. So what ever number you got for b/2 will be in your perfect square binomial.  Moving on to solve from here we need to FIRST undo the square by taking the square root of both sides and then adding the nine. So our final answer is the positive and negative value of the square root of 30 plus 9. I show step by step how to solve problems just like this on my youtube blog at www.youtube.com/mrbrianmclogan
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how do i solve the quadratic equation 16x^2=25

This is a perfect square.  First set it equal to 0 by subtracting 25 from both sides 16x^2-25=0 Take the Square root of both terms Square root of 16x^2 is 4x Square root of -25 is 5 and -5 The factored form is (4x+5) (4x-5) If you set each of these equal to 0 and solve for x you get 4x+5=0 subtract 5 4x=-5 divide by 4 -5/4=x Do the same with the next factor 4x-5=0 add 5 4x=5 divide by 4 x=5/4 the two roots for this are 5/4 and -5/4
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how to solve yhis in equation cube square root of negative 25

????? "square kube root ???? square root (-25)=5i weer big krats define i=root(-1)
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show that the function f(x)= sqrt (x^2 +1) satisfies the 2 hypotheses of the Mean Value Theorem

f(x) = sqrt(x^2 + 1) ; [(0, sqrt(8)] Okay, so for the Mean Value Theorem, two things have to be true: f(x) has to be continuous on the interval [0, sqrt(8)] and f(x) has to be differentiable on the interval (0, sqrt(8)). First find where sqrt(x^2 + 1) is continous on. We know that for square roots, the number has to be greater than or equal to zero (definitely no negative numbers). So set the inside greater than or equal to zero and solve for x. You'll get an imaginary number because when you move 1 to the other side, it'll be negative. So, this means that the number inside the square root will always be positive, which makes sense because the x is squared and you're adding 1 to it, not subtracting. There would be no way to get a negative number under the square root in this situation. Therefore, since f(x) is continuos everywhere, (-infinity, infinity), then f(x) is continuous on [0, sqrt(8)]. Now you have to check if it is differentiable on that interval. To check this, you basically do the same but with the derivative of the function. f'(x) = (1/2)(x^2+1)^(-1/2)x2x which equals to f'(x) = x/sqrt(x^2+1). So for the derivative of f, you have a square root on the bottom, but notice that the denominator is exactly the same as the original function. Since we can't have the denominator equal to zero, we set the denominator equal to zero and solve to find the value of x that will make it equal to zero. However, just like in the first one, it will never reach zero because of the x^2 and +1. Now you know that f'(x) is continous everywhere so f(x) is differentiable everywhere. Therefore, since f(x) is differentiable everywhere (-infinity, infinity), then it is differentiable on (0, sqrt(8)). So the function satisfies the two hypotheses of the Mean Value Theorem. You definitely wouldn't have to write this long for a test or homework; its probably one or two lines of explanation at most. But I hope this is understandable enough to apply to other similar questions!
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how would a quadratic problem look

a quadratic problem is really only a way to solve a certain polynomial. if i gave you the equation: x(squared) + 4x + 4 then you can solve it using the quadratic formula which is: (b (plus or minus) ( the square root of ) -b(squared) - 4ac )/ 2 the 'a'b'and c' in the equation are these: ax(squared) + bx + c a=1 , b=4, and c=4 so (4(plus or minus the square root of) -4(squared) - 4(1)(4))/2 4 (plus or minus the square root of) 16-16/2 16 - 16 =0 and 4/2 is 2
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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