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How to solve a quadratic equation? step by step

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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 ...

Solving Quadratic Equations by Factoring | Purplemath

solve, solving, quadratic, quadratics, equation, equations, Quadratic Formula, factor, factoring, square, root, zero, product, property, solution, Purplemath

The Quadratic Formula Explained | Purplemath

The Quadratic Formula Explained. ... and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve. ...

How to Solve Quadratic Equations - Algebra-Class.com

Feb 01, 2011 · ... shows how to use the quadratic formula to solve quadratic equations. ... Solving Quadratic Equations: The Quadratic Formula [fbt] - Duration: 21:04.

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. Just enter the values of a, b and c below

Use the quadratic formula to solve the equation, 0 is equal to negative 7q squared plus 2q plus 9. Now, the quadratic formula, it applies to any quadratic equation of ...

︎² How to Solve Quadratic Equations by Factoring (mathbff ...

Aug 18, 2013 · MIT grad shows how to solve any quadratic equation by factoring. To skip to the shortcut trick, go to time 6:11. More videos with Nancy coming in 2017! The ...

Solve a Quadratic Equation by Factoring - WebMath

This page will try to solve a quadratic equation by factoring it first. How does this work? Well, suppose you have a quadratic equation that can be factored, like

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.

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

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

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 =

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.

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)

If a quadratic equation can be solved, how many solutions are there?

Most of the quadratic eqation have 1 or 2 solution Most solveable quadratics have 2 distinct roots. x² + 5x + 4 = 0 (x + 4)(x + 1) = 0 Roots are -4 and -1 There is only one solution if the quadratic has a double root of the same value. x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 x = -3 is the one root. Understand more about Quadratic Equation

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.

Quadratic Functions, Dividing Polynomials, and Zeros of Polynomials

1. The maximum area is when the rectangle is a square. So the three fenced sides are equal and have length 150/3=50 yd. Let's look at the logic: the sides of the rectangle are L and W so the area, A, is LW and the perimeter, P=2L+2W. L=(P-2W)/2 and A=W(P-2W)/2=(1/2)(WP-2W^2). The maximum value of A is obtained by differentiating the quadratic with respect to W: (1/2)(P-4W)=0 at a turning point, so W=P/4 and L=(P-P/2)/2=P/4. Therefore L=W=P/4. The enclosed area is therefore a square, and we know that the three sides plus another unfenced length, L, make up the perimeter P=4L=150+L, so 3L=150 and L=50 yd. Alternative solution avoiding calculus: a. A=LW where L=length of rectangular area, W=width. b. 150=2L+W. c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L). d. A=L(P-2L)/2=(LP-2L^2)/2=LP/2-L^2. e. f(L)=-L^2+LP/2-A in standard quadratic form f(x)=ax^2+bx+c, where a=-1, b=P/2, c=-A and x=L. f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area. g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16. From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd. h. Maximum area is 50^2=2500 sq yd. 2. Synthetic division (1) -3 | 2 -3 -45 -54 ......2 -6..27...54 ......2 -9 -18 | 0 3. Synthetic division (2) 6 | 2..-9 -18 .....2 12..18 .....2...3..| 0 This quotient is 2x+3, representing the third factor of the polynomial. 4. Synthetic division (3) -4 | 2 -3 -45..-54 ......2 -8..44.....4 ......2 -11 -1 | -50 The Remainder Theorem tells us that substituting x=-4 into the polynomial gives us the same remainder. The substitution gives: -128-48+180-54=-50. The zeroes are -3, 6 and -3/2, because synthetic division (2) tells us that 2x+3 is a factor.

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