Guide :

How do you solve for (X) in this quadratic equation, 3X^2/(X^2-1)=4/(X-1)

solve by the use of factorization

Research, Knowledge and Information :


Quadratic Equation Solver - Math is Fun


Quadratic Equation Solver. If you have an equation of the form "ax 2 + bx + c = 0", we can solve it for you. ... x 2 = 3x -1: x 2 - 3x + 1 = 0: a=1, ...
Read More At : www.mathsisfun.com...

3 Ways to Solve Quadratic Equations - wikiHow


How to Solve Quadratic Equations. A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. There are three ...
Read More At : www.wikihow.com...

How do I solve [math]4^x-2^x-2=0[/math] - Quora - A place to ...


... \Leftrightarrow y^2-y-2=0[/math] … omitting quadratic equation resolution ... How do I solve [math]x^6-x^5+3x^4-x^2-3=0 ... How do you solve the problem 4^x = 1/2?
Read More At : www.quora.com...

How do you solve x^2+3x+2=0? | Socratic


x=-2 or x=-1 Two standard ways to solve a quadratic equation: Firstly you could factorise it to the form:- x^2+3x+2=0 x^2+(a+b) ... How do you solve #3x^2 = 108#?
Read More At : socratic.org...

How do you solve x^2 + 3x + 6 = 0 by quadratic formula ...


SOCRATIC Subjects . Science Anatomy ... How do you solve #x^2 + 3x + 6 = 0# by quadratic formula? ... #x^2 +3x +6 =0# The equation is of the form #color ...
Read More At : socratic.org...

How can the quadratic equation [math]x^2 - 3x - 10 = 0[/math ...


How can the quadratic equation [math]x^2 - 3x - 10 = 0[/math] be solved? Update Cancel. Answer Wiki. ... How do you solve 3x^2-2x-3=0 with the quadratic formula?
Read More At : www.quora.com...

Algebra 2 - Solve by using the quadratic formula when c=0, 3x ...


Feb 26, 2013 · http://www.freemathvideos.com in this video tutorial I show you how to solve a quadratic equation by using the quadratic equation. When factoring or ...

Solve The Quadratic Equation X^2- 3x- 2 = 0


Here you find answers to questions like: Solve The Quadratic Equation X^2- 3x- 2 = 0 . Or how to solve the quadratic equation where the coeficients are a = 1, b = -3 ...
Read More At : coolconversion.com...

Solving Quadratic Equations by Factoring | Purplemath


The new thing here is that the quadratic expression is part of an equation, and you're told to solve for the values of the ... when solving quadratic equations ...
Read More At : www.purplemath.com...

Suggested Questions And Answer :


2/×+2+4/×-5=28/(×+2)(×-5)

If you mean 2/(x+4) + 4/(x-5) = 28/( (x+2)(x-5) ) then: Multiply both sides by (x+4)(x+2)(x-5) 2(x+2)(x-5) + 4(x+4)(x+2) = 28(x+4) x(x^2 - 3x - 10) + 4(x^2 + 6x + 8) = 28x + 112 x^3 - 3x^2 - 10x + 4x^2 + 24x + 32 = 28x + 112 x^3 + x^2 + 14x + 32 = 28x + 112 x^3 + x^2 - 14x - 80 = 0 Checking for nice solutions. . . 80: 2 * 2 * 2 * 2 * 5 What you can make with the prime factors of 80:  2, 4, 5, 8, 10, and a bunch of larger numbers. If you plug 10 in for x, the x^3 makes 1000, much larger than the rest of the equation, so 10 is not a root of x^3 + x^2 - 14x - 80 = 0 2: -96 4: -56 5: 125 + 25 - 70 - 80 = 0 x = 5 is a root (x - 5)( ? ) = x^3 + x^2 - 14x - 80 (x-5)(x^2 + ?x + 16) = x^3 + x^2 - 14x - 80 The -14x is made of 16x + (-5)(?x) -14x = 16x -5?x -14 = 16 - 5? -30 = -5? ? = 6 (x-5)(x^2 + 6x + 16) = x^3 + x^2 - 14x - 80 Checking (x-5)(x^2 + 6x + 16). . . x^3 + 6x^2 + 16x -5x^2 - 30x - 80 x^3 + x^2 - 14x - 80  good (x-5)(x^2 + 6x + 16) factor x^2 + 6x + 16 If you're in pre-algebra, you won't have run into it yet, but there's this thing called the quadratic formula for solving and factoring things with x^2 If you have ax^2 + bx + c = 0 then the values for x are x = (-b +- sqrt(b^2 - 4ac) ) / 2a x^3 + x^2 - 14x - 80 = (x-5)(x^2 + 6x + 16) = 0 a = 1, b = 6, c = 16 x = (-6 +- sqrt(6^2 - 4(1)(16)) ) / 2(1) x = (-6 +- sqrt(36 - 64) ) / 2 x = (-6 +- sqrt(-28)) / 2 You can't take the square root of a negative number, so doesn't factor. This means there is no way to make x^2 + 6x + 16 = 0. That means the only way to make (x-5)(x^2 + 6x + 16) = 0 is if x - 5 = 0, which gives us x = 5 Answer:  x = 5
Read More: ...

Prove the quadratic equation

A quadratic equation in the standard form is given by ax 2 + bx + c = 0 where a, b and c are constants with a not equal to zero. ...Solve the above equation to find the quadratic fomulas Given ax 2 + bx + c = 0 Divide all terms by a x 2 + (b / a) x + c / a = 0 Subtract c / a from both sides x 2 + (b / a) x + c / a - c / a = - c / a and simplify x 2 + (b / a) x = - c / a Add (b / 2a) 2 to both sides x 2 + (b / a) x + (b / 2a) 2 = - c / a + (b / 2a) 2 to complete the square [ x + (b / 2a) ] 2 = - c / a + (b / 2a) 2 Group the two terms on the right side of the equation [ x + (b / 2a) ] 2 = [ b 2 - 4a c ] / ( 4a2 ) Solve by taking the square root x + (b / 2a) = ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } Solve for x to obtain two solutions x = - b / 2a ± sqrt { [ b 2 - 4a c ] / ( 4a2 ) } The term sqrt { [ b 2 - 4a c ] / ( 4a2 ) } may be written sqrt { [ b 2 - 4a c ] / ( 4a2 ) } = sqrt(b 2 - 4a c) / 2 | a | Since 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written x = [ -b + sqrt( b 2 - 4a c ) ] / 2 a x = [ -b - sqrt( b 2 - 4a c ) ] / 2 a The term b 2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible case 1: If b 2 - 4a c > 0 , the equation has 2 solutions. case 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. case 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions.
Read More: ...

solve for x when (.182+x)/(.106-2x)=3.24

solve for x when (.182+x)/(.106-2x)=3.24 Im trying to solve for x its for a chemistry equation but i need to solve for x using the quadratic equation and i dont know which equals a b and c to find x. ************************** You can't solve this with the quadratic formula, because there is no x^2 term. Observe: (.182+x)/(.106-2x)=3.24 Multiply both sides by (.106-2x) to eliminate the fraction on the left. (.106-2x) * (.182+x)/(.106-2x) = (.106-2x) * 3.24 0.182 + x = 0.34344 - 6.48x This is a linear equation. Even if you arranged it so that all the terms are on the left, leaving a zero on the right side, you still would not have an x^2 term. That means that the "a" coefficient would be zero. When you try to use that in the quadratic formula, the denominator would be 2a = 2 * 0 = 0. You cannot divide by zero, so you are now stuck, with no possible solution. If you solve the equation from where I have taken it, you will find the value of x. 0.182 + x = 0.34344 - 6.48x 6.48x + x = 0.34344 - 0.182 7.48x = 0.16144 (Note that you could now subtract 0.16144 from both sides to get 7.48x - 0.16144 = 0, but you still don't have an x^2 term, so a=0, b=7.48 and c=-0.16144. That pesky 0 for the value of a will give that 0 denominator. Impossible to divide.) x = 0.0215828877
Read More: ...

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.
Read More: ...

solve for x in this quadratic equation: 4x^2 + 5x = 6

Yes, sometimes factoring doesn't work. That's when we fall back on the quadratic formula. 4x^2 + 5x - 6 = 0 This is in the general form, y = ax^2 + bx + c. From that, we extract the values we need when we use the quadratic formula.        -b ± sqrt(b^2 - 4ac) x = --------------------------                2a        -5 ± sqrt(5^2 - 4(4)(-6)) x = -------------------------------                  2(4)        -5 ± sqrt(25 + 96) x = ------------------------                 8        -5 ± sqrt(121) x = --------------------               8        -5 ± 11 x = -----------           8        -5 + 11                       -5 - 11 x = -----------     and    x = -----------            8                                8        6                      -16 x = ---     and     x = ------       8                         8        3 x = ---     and     x = -2       4 Always check your answers. 4x^2 + 5x = 6 4(3/4)^2 + 5(3/4) = 6 4(9/16) + 15/4 = 6 9/4 + 15/4 = 6 24/4 = 6 6 = 6             That one checks. 4x^2 + 5x = 6 4(-2)^2 + 5(-2) = 6 4(4) - 10 = 6 16 - 10 = 6 6 = 6            That one checks, too. Answer: x = 3/4   and x = -2 The question was: solve for x in this quadratic equation: 4x^2 + 5x = 6 in this quadratic equation 4x^2 + 5x -6 = 0 how can I solve for x? How does it go step by step. I tried factoring, but got stuck because nothing made sense (2x +    )   (2x   -    )  = 0
Read More: ...

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer

3x + 6y – 6z = 9 2x – 5y + 4z = 6 -x +16y + 14z = -3 what is the answer how do you solve it and what are the answers ? 1)  3x + 6y – 6z = 9 2)  2x – 5y + 4z = 6 3)  -x +16y + 14z = -3 The first objective is to eliminate z so we can solve for x and y. It's a multi-step process, so follow along. Multiply equation one by 4. 4 * (3x + 6y – 6z) = 9 * 4 4)  12x + 24y - 24z = 36 Multiply equation 2 by 6. 6 * (2x – 5y + 4z) = 6 * 6 5)  12x - 30y + 24z = 36 Now, we have two equations with a "24z" term. Add the equations and the z drops out. Add equation five to equation four.    12x + 24y - 24z = 36 +(12x - 30y + 24z = 36) ----------------------------------    24x -  6y           = 72 6)  24x - 6y = 72 The same process applies to equations two and three. Multiply equation two by 7 this time. 7 * (2x – 5y + 4z) = 6 * 7 7)  14x - 35y + 28z = 42 Multiply equation three by 2. 2 * (-x + 16y + 14z) = -3 * 2 8)  -2x + 32y + 28z = -6 Subtract equation eight from equation seven.   14x - 35y + 28z = 42 -(-2x + 32y + 28z = -6) ---------------------------------   16x - 67y          = 48 9)  16x - 67y = 48 Looking at equations six and nine, it would be simpler to eliminate the x. The multipliers are smaller. Multiply equation six by 2. 2 * (24x - 6y) = 72 * 2 10)  48x - 12y = 144 Multiply equation nine by 3. 3 * (16x - 67y) = 48 * 3 11)  48x - 201y = 144 Subtract equation eleven from equation 10.   48x -   12y = 144 -(48x - 201y = 144) ---------------------------          189y  =   0 189y = 0 y = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Substitute that value into equations six and nine to solve for x and verify. Six: 24x - 6y = 72 24x - 6(0) = 72 24x - 0 = 72 24x = 72 x = 3  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Nine: 16x - 67y = 48 16x - 67(0) = 48 16x - 0 = 48 16x = 48 x = 3    same answer for x To solve for z,substitute the x and y values into the three original equations. One: 3x + 6y – 6z = 9 3(3) + 6(0) – 6z = 9 9 + 0 - 6z = 9 9 - 6z = 9 -6z = 9 - 9 -6z = 0 z = 0  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<< Two: 2x – 5y + 4z = 6 2(3) – 5(0) + 4z = 6 6 - 0 + 4z = 6 6 + 4z = 6 4z = 6 - 6 4z = 0 z = 0     same answer Three: -x +16y + 14z = -3 -(3) +16(0) + 14z = -3 -3 + 0 + 14z = -3 -3 + 14z = -3 14z = -3 + 3 14z = 0 z = 0     once again, same answer x = 3, y = 0, z = 0
Read More: ...

Find the orthogonal canonical reduction of the quadratic form −x^2 +y^2 +z^2 −6xy−6xz+2yz. Also, find its principal axes, rank and signature of the quadratic form.

4.4 Systems of Equations - Three Variables Objective: Solve systems of equations with three variables using addi- tion/elimination. Solving systems of equations with 3 variables is very simila r to how we solve sys- tems with two variables. When we had two variables we reduced the system down to one with only one variable (by substitution or addition). With three variables we will reduce the system down to one with two variables (usua lly by addition), which we can then solve by either addition or substitution. To reduce from three variables down to two it is very importan t to keep the work organized. We will use addition with two equations to elimin ate one variable. This new equation we will call (A). Then we will use a different pair of equations and use addition to eliminate the same variable. This second new equation we will call (B). Once we have done this we will have two equation s (A) and (B) with the same two variables that we can solve using either met hod. This is shown in the following examples. Example 1. 3 x + 2 y − z = − 1 − 2 x − 2 y + 3 z = 5 We will eliminate y using two different pairs of equations 5 x + 2 y − z =
Read More: ...

please help solve this quadartic problem step by step 26r-2=3rto the 2nd power also to simlify it in the end

26r-2=3r^2. The standard quadratic form is ax^2+bx+c=0, where x is the unknown and a, b and c are numbers. The unknown is r in this problem, so let's put the equation into quadratic form: 3r^2-26r+2=0. Note the sign changes as terms are transferred from one side to another, and also note that, in order to preserve the 3r^2 on the right I've brought all the terms over to the right to bring them all together. In doing so, which would have made the equation 0=3r^2-26r+2, I've moved the equals and zero over to the right to get it into quadratic form. I can do this because if A=B then it's true that B=A. This equation doesn't factorise, so we need the formula to solve it: r=(26+sqrt(26^2-4*3*2))/6 (this is the quadratic formula x=(-b+sqrt(b^2-4ac))/2a). So r=(26+sqrt(676-24))/6=(26+sqrt(652))/6=8.589 or 0.0776.
Read More: ...

solve 14x-y=6 for y

14x-y=6 Since 14x does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 14x from both sides. -y=-14x+6 Multiply each term in the equation by -1. -y*-1=-14x*-1+6*-1 Multiply -y by -1 to get y. y=-14x*-1+6*-1 Simplify the right-hand side of the equation by multiplying out all the terms. y=14x-6
Read More: ...

Solve the equation p^2+p(x+y)+xy=0

Question:  Solve the equation p^2+p(x+y)+xy=0. Its related to engineering math -  Clairaut's equation. Since Clairaut's equation is a differential equation, then I am assuming that p means the first differential, dy/dx, or y'. You equation actually then is: (y')^2 + (y')*(x+y) + xy = 0. Treating your equation as a simple quadratic equation, and using the quadratic formula to solve it, p = {-(x+y) +/- sqrt((x+y)^2 - 4xy)}/(2*1) p = {-(x+y) +/- sqrt((x-y)^2)}/2 p = {-(x+y) +/- (x-y)}/2 y' = {-(x+y) - (x-y)}/2,  y' = {-(x+y) + (x-y)}/2, y' = -x,   y' = -y And from these the solutions are, y1(x) = C - (1/2)x^2,  y2(x) = ke^(-x)
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.1426 seconds