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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.

this question is from Linear Algebra.

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Find the orthogonal canonical reduction of the quadratic from ...


Find the orthogonal canonical reduction of the quadratic from -x^2+y^2+z^2-6xy+2yz. Also, find its principal axes,rank ... reduction of the quadratic from -x^2+y^2+z ...
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Reduce the quadratic form Q(x)=x^2+y^2+z^2+2xy-4yz+6xz to ...


Reduce the quadratic form Q(x)=x^2+y^2+z^2 ... Find the orthogonal canonical reduction of the quadratic from -x^2+y^2+z^2-6xy+2yz. Also, find its principal axes,rank ...
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MTE-02 LINEAR ALGEBRA (1 December, 2016)


... Find the orthogonal canonical reduction of the quadratic form x2 +y2 +z2 6xy 6xz+2yz. Also, find its principal axes, rank and signature of the quadratic form ...
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9.1 matrix of a quad form - U.I.U.C. Math - Illinois


Multiply out and collect terms to get q in terms of X and Y: q = 17X 2 +6XY ... 2 What is the matrix of a quadratic form. ... X,Y coord system which has the same axes ...
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MTE -02 ASSIGNMENT BOOKLET Bachelor’s Degree Programme (B.Sc ...


ASSIGNMENT BOOKLET Bachelor’s Degree ... Find the orthogonal canonical reduction of the quadratic form x2 2y2 +z2 +2xy+6yz and its principal axes. Also, find the ...
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Quadratic Polynomials - PDF - docplayer.net


... the next step is to study quadratic polynomials, y = ax 2 ... compact form where X = ( ) x y Similarly, we can also write ... a) q(x, y) = x 2 6xy + 9y 2. (b) Q(x ...
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Newest Linked Questions - Mathematics Stack Exchange


Linked Questions. hot newest votes active ... a question about a canonical form of a quadratic form using ... =x_1^{2}+x_3^2+4x_1x_2-4x_1x_3 $$ to obtain the ...
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Engineering Mathematics - I Question Bank - Docs.com


Reduce the quadratic form 2 2 2 2 5 3 4 x y z xy ... form by an orthogonal reduction. Also find its ... Also find the rank, index, signature and ...
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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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Finding the nontrivial perfect square of a given quadratic form?

assuming z^2+180-88=m^2 rewrite it in a form of z(z+180) = m^2 + 88 since LHS is a form of 2 factors z and z+180, the right handside must be also in such a form as a x b therefore m must have same factors as 88, you can start from there and try the factors of 88
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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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volume of a solid bounded by the gragh of the equation √(x/a)+√(y/b)+√(y/c) =1 and the coordinate planes

I read the equation as √(x/a)+√(y/b)+√(z/c)=1 √(z/c)=1-√(y/b)-√(x/a) and z/c=(1-√(y/b)-√(x/a))^2. If we consider a cuboid (rectangular prism) of height z and length and width dx and dy, then the volume of the cuboid is zdxdy. The double integral will be ∫∫zdxdxy, representing the sum of an infinite number of cuboids with an infinitesimal cross-section, is the volume of the required solid. In this double integral we replace z by c((1-√(y/b)-√(x/a))^2). Now we need to work out the limits. The positive square roots are implied. Put z=0 into the equation. What we get is the x-y plane and a view of the end face of the solid. The curve meets the y-axis at (0,b) and the y-axis at (a,0). These are the y and x intercepts. Therefore we have the x and y limits for the double integral. For z>0 (up to z=c) we have laminae building up from the x-y plane. The area of one infinitesimally thin lamina is the area under the curve bounded by the intercepts. The sum of the laminae is the volume of the required solid. When z=c we reach the limit of the solid. The question does not ask for evaluation of the double integral, but the process is to integrate c((1-√(y/b)-√(x/a))^2) wrt x treating y as a constant, between the limits x=0 and x=a; then the result is integrated wrt y between the limits y=0 and y=b. The limit of z=c isn't used because the equation defining z has been used as the integrand. The answer is c∫∫((1-√(y/b)-√(x/a))^2)dxdxy with 0≤x≤a and 0≤y≤b. The picture below shows in "net" form the general shape of the solid as folded-down axes to produce a 2-dimensional impression. By folding up the axes so that they form a 3-dimensional set of axes it's possible to imagine what the solid looks like: With Z-X as the base plane fold up Y-Z and X-Y so that the markers -2 and -1 in Y-Z are "stitched" to the markers 2 and 1 in X-Y  making the X, Y and Z axes orthogonal like the corner of a cube. The result is a sort of cobweb strung across the corner, where the curves are the strands of the cobweb fastened to the "walls". Now mentally fill in the body of the cobweb, which is the required solid.
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how i solve this f(x)=0.3582*x^2 - 3.3833 * x + 9.0748 by matlab

how i solve this f(x)=0.3582*x^2 - 3.3833 * x + 9.0748 When asked to solve f(x) = 0, that means find the value(s) of x that, when substituted into the expression for f(x) will make it equal to zero. So, to find f(x) = 0, then we set 0.3582*x^2 - 3.3833 * x + 9.0748 = 0. This is a quadratic equation, of the form ax^2 + bx + c = 0, with a = 0.3582, b = - 3.3833 and c = 9.0748. We use the quadratic formula to solve quadratic equations. This is, x = {-b +/- sqrt(b^2 - 4ac)} / (2a) substituting for the values given for a, b and c, x = {3.3833 +/- sqrt((-3.3833)^2 - 4*(0.3582)*(9.0748))} / (2*0.3582) x = {3.3833 +/- sqrt(11.4467 - 13.0024)} / (0.7164) x = {3.3833 +/- sqrt(-1.55565)} / (0.7164) Since the discriminant is negative (-1.55565), that means that we have no real values for x. The expression, 0.3582*x^2 - 3.3833 * x + 9.0748, is a polynomial of 2nd degree. If you were to plot this curve you would get a parabola (U-shaped upwards) and its vertex would be above the x-axis. That means the curve never crosses the x-axis, which is why there is no (real) solution. Our solution(s) then are complex numbers, made up of real and imaginary parts. We write the square root part as:  sqrt(-1.55565) = sqrt(1.55565) * sqrt(-1) = sqrt(1.55565)}*i, where i = sqrt(-1). So now we have, x = {3.3833 +/- sqrt(1.55565)*i} / (0.7164) x = {3.3833 +/- 1.24726*i} / (0.7164) x = 4.72264 +/- 1.741*i
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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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y=-x^2+35x

We find the x-intercepts by setting y to zero. -x^2 + 35x = 0 x^2 - 35x = 0 We use "completing the square." x^2 - 35x + (35/2)^2 = (35/2)^2 x^2 - 35x + 17.5^2 = 17.5^2 (x - 17.5)^2 = 17.5^2 x - 17.5 = ±17.5 x = 17.5 ±17.5 x = 17.5 + 17.5    and    x = 17.5 - 17.5 x = 35   and    x = 0 The x-intercepts are (0, 0)  and (35, 0) The vertex has an x value halfway between them: 17.5 We can calculate the y value of the vertex using x=17.5 y = -x^2 + 35x y = -(17.5)^2 + 35(17.5) y = -306.25 + 612.5 y = 306.25 The vertex is at (17.5, 306.25) Graphing this equation would show a parabola opening downward, with the highest point at (17.5, 306.25), crossing the x-axis at (0, 0) and (35, 0).   Problem statement: y=-x^2+35x I need help on 4 quadratics and turn them into graghs with a form of a parbola
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Equations of a line, application of maxima and minima.

There might be an easier way but here's one: The area of the triangle is : A = 0.5(x0)(y0), where x0 and y0 are the x and y intercepts. We want to minimise this. The equation for a line is y = mx + y0, where m is the gradient and y0 is the y intercept. Using (4,3) coordinate, we get 4 = 3m + y0, or y0 = 4 - 3m now we need to find the x intercept in terms of m. This happens when y = 0: 0 = m(x0) + y0 = m(x0) + (4 - 3m) Rearranging gives: x0 = 3 - 4/m Now we can work out the area: A = 0.5 (x0) (y0) = 0.5 (3-4/m) (4-3m) A = 0.5 (12 - 9m - 16/m + 12) = 0.5 (24 - 9m - 16/m) To minimise the area, we can differentiate it with respect to m and let that equal 0: dA/dm = d[ 0.5 ( 24 - 9m - 16/m) ] = 0 -9 + 32/m^2 = 0 m = +/- sqrt(32/9) Because the line intercepts both x and y axes in the positive section, the gradient, m, has to be negative. Substituting to find the y intercept to complete the equation: c = 4 - 3m = 4 + 3(sqrt(32/9)) So the final answer is: y = 4 + 3(sqrt(32/9)) - sqrt(32\9)x or y = 9.657 - 1.886x
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Can someone help solve this in quadratic formula, step by step?4/9w2+1=5/3w

4/9w^2 +1 = 5/3w        using completing the square method,first re arrange the equation                         4/9w^2 -5/3w +1 = 0      to get a whole number, multiply  through by 9         4w^2-15w+9=0      next,divide through by the co efficient of w^2     w^2-15/4w+9/4=0      take the constant to the other side      4w^2-15w=-9/4     next, find half of the co efficient of w and add it to both sides                                 w^2-15/4w+(15/8)^2 = (15/8)^2 - 9/4      since the left hand side is a perfect square, it becomes  (w+15/8)^2 = (15/8)^2 -9/4.       (w+15/8)^2 = 225/64-9/4. (find the lcm of the right hand side )       now square both sides         (w+15/8) = squareroot 81/64(squareroot =+_ 9/8)      collecting like terms,    w= 15/8 +_9/8       w=15+9/8 or w=15-9/8      :.w is either 3 or3/4
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Find the vertex of h=-0.8(t-4)^2+12.8

The function is a parabola in inverted U form. The form of the equation tells us that the vertex is at t=4, h=12.8, because this point is uniquely the point where there is only one value of t, instead of two, for each value of h in range. The vertex at (4,12.8) sits between the zeroes of the function given by putting h=0. 0.8(t-4)^2=12.8, so (t-4)^2=16 and t-4=±4. The zeroes are at (0,0) (the origin, where the h (vertical) and t (horizontal) axes intersect) and (8,0). This information should be sufficient to sketch the graph. We can tabulate some values: t h -2 -16 0 0 2 9.6 4 12.8 6 9.6 8 0 10 -16
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