Guide :

solve this equation as an reducible to the quadratic

(2x-4)^2-4(2x-4)-5=0

Research, Knowledge and Information :


Algebra - Equations Reducible to Quadratic in Form


... (Notes) / Solving Equations and Inequalities / Equations Reducible to Quadratic in Form. Algebra ... is a quadratic equation and we can solve that.
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Equations which are Reducible to Quadratic - Tiger Algebra


Equations which are Reducible to Quadratic. The Tiger Algebra solver shows you, ... Solving Linear Equations by Substitution; Properties of a Straight Line;
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10.4 Solving Equations in Quadratic Form, Equations Reducible ...


10.4 Solving Equations in Quadratic Form, ... Now that we can solve all quadratic equations we want to solve ... Notice that the quadratic equations in our examples ...
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Equations Reducible to Quadratic Form, Quadratic Equation ...


Math Algebra 1 Quadratic Equation Equations Reducible to Quadratic Form. Top. Equations Reducible to Quadratic Form. ... Just recall that for solving a quadratic ...
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Equations Reducible to Quadratic Form - OoCities


Equations Reducible to Quadratic Form ... The equation contains only one radical. Solve . ... substitution in order to rewrite the equation in quadratic form. Solve .
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OpenAlgebra.com: Solving Equations Quadratic in Form


Solving Equations Quadratic in ... here are reducible to quadratic ... used to easily solve many other equations that are quadratic in form. Solve by making a ...
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Equations Reducible to Quadratic Equations Exercise 4


Equations Reducible to Quadratic Equations Exercise 4.2 Solve the following equations: 1. x x4 2− + =6 8 0 Solution: ... Using Quadratic formula: ...
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Equations reducible to quadratic form - Ashworth College


Equations reducible to quadratic form ... x2 −3x +2 =0 Now we solve this equation. I will use factoring ()x −2 (x −1) ...
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Algebra 2 Section 8.5 Equations Reducible to Quadratic Form


Feb 11, 2013 · Algebra 2 Section 8.5 Equations Reducible to Quadratic Form Kristin Gilchrist. Loading... ... Solving Quadratic Equations by Factoring ...

Errors In Solving Equations Reducible To Quadratic Form ...


... solving equations reducible into quadratic ... This present study is an attempt to examine students' errors in solving equations reducible to quadratic ...
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Suggested Questions And Answer :


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 =
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solving a quadratic equation

solving a quadratic equation 2x²-4x+3=0 We solve for possible values of x using the following formula:        -b ± sqrt(b^2 - 4(a)(c)) x = --------------------------------                2(a)        -(-4) ± sqrt((-4)^2 - 4(2)(3)) x = --------------------------------------                2(2)        4 ± sqrt(16 - 24) x = --------------------------------------                4        4 ± sqrt(-8) x = --------------------------------------                4 We can stop right there. This part of the formula, sqrt(b^2 - 4(a)(c)), is called the discriminant. It will give us a clue as to how many roots there are for the equation. As you can see above, it reduced to a negative number inside the square root operation. This gives us imaginary numbers. Having a negative discriminant means there are no values of x where the graph of this equation crosses the x-axis.
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zeros of 2x^6+5x^4-x²

All the powers of 2 are even so this 6-degree expression can be reduced to a cubic if we put y=x^2: 2y^3+5y^2-y. Putting this equal to zero to find the y zeroes: 2y^3+5y^2-y=0. This factorises: y(2y^2+5y-1)=0. Take the quadratic: 2y^2+5y-1=0 and rewrite it as: y^2+5y/2=1/2 by dividing through by 2 and moving the constant across. Now we can complete the square by dividing the y term by 2 and squaring its coefficient: y^2+5y/2+25/16=1/2+25/16, and you can see we've added 25/16 to both sides of the equation to make it balance. We now have a perfect square on the left: (y+5/4)^2=1/2+25/16=(8+25)/16=33/16, so we can take the square root of each side:  (y+5/4)=+sqrt(33)/4. Therefore, y=-5/4+sqrt(33)/4. This is the same result as we would have found using the formula for solving a quadratic equation. We need to calculate what these solutions are: 0.18614 and -2.68614. But we want x, not y, so we need to find the square root of these values of y. Assuming we don't want complex solutions (involving the imaginary square root of -1), we can only use the positive solution 0.18614 and take the square root of that, which is +0.43144 approx. so the real solution is x=0.43144 or -0.43144. (The complex solution is +1.63894i, where i=sqrt(-1).) But we're not finished yet, because y=0 was also a solution, and that means x=0 is a solution, so we have three possible real zeroes for x: 0, 0.43144 and -0.43144 (and two complex zeroes).
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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.
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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
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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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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
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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.
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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)
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The pythagorean theorem of this problem x^2+(2x+6)^2=(2x+4)^2

Pythagoras' theorem relates the lengths of the sides of a right-angled triangle: a^2+b^2=c^2 where c, the hypotenuse, is the longest side. In your question 2x+6 is the longest side, so there could be no solution, since the the sum of the two squares of the sides on the left-hand side must be bigger than the square of a side that's smaller than either one of them, and 2x+4 is smaller than 2x+6 (2x+6-(2x+4)=2). The equation x^2+(2x+4)^2=(2x+6)^2 can be solved. We can write it as: x^2=(2x+6)^2-(2x+4)^2. On the right-hand side we have the difference of two squares, which factorises to make the calculation simpler: (2x+6-(2x+4))(2x+6+2x+4)=2(4x+10). So now we have x^2=8x+20, by expanding the brackets. We can write this: x^2-8x=20 by subtracting 8x from each side. This is a quadratic that can be solved by completing the square. Halve the x coefficient then square it: 8/2=4, and 4^2=16. Add 16 to both sides: x^2-8x+16=36. The left-hand side is a perfect square of x-4: (x-4)^2=36. 36 is also a perfect square=6^2 so, by taking square roots of each side we have: x-4=6 or -6, because both 6 and -6 have a square of 36. We have two solutions: x=4+6=10 or 4-6=-2. Another way of solving for x is factorisation: x^2-8x=20 can also be written: x^2-8x-20=0 which factorises: (x-10)(x+2)=0, so either x-10=0 or x+2=0, and that also gives us x=10 and -2. The factors x-10 and x+2 are binomial expressions, and the equations arising from them use the zero factor idea: x-10=0 or x+2=0 to solve for x.
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