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# how to solve quadratic expression?

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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 ...

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

Aug 18, 2013 · MIT grad shows how to solve any quadratic equation by factoring. ... ︎² How to Solve Quadratic Equations by Factoring (mathbff) mathbff. Loading ...

### Quadratic Equations - Math is Fun

Students learned to solve quadratic equations by ... Being able to use a graphing calculator to solve a quadratic equation requires the ability to produce a ...

### Quadratic Formula Calculator and Solver will calculate ...

Quadratic Formula Calculator & Solver. ... but if you have access to a graphing calculator you should be able to solve quadratic equations, ...

Solving quadratics by factoring: ... Prepare with these 10 lessons on Quadratic equations and functions. ... Solving quadratic equations ...

### Solving Quadratic Equations - San Jacinto College

Solving Quadratic Equations Terminology. 1. ... Steps for solving Quadratic application problems: 1. Draw and label a picture if necessary. 2.

### 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

## Suggested Questions And Answer :

### 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

### please solve the rational expression 6y^2-52y-18/3y^2+25y+8*3y^2+18y-48/y^2-11y+18

Could you make sure that you include brackets in your expresion(s). It's ambiguous otherwise, and sometimes expressions like this are impossible to answer. I'm having to guess at what I think the (rational) expression is supposed to be. Thanks. solve: (6y^2-52y-18)/(3y^2+25y+8)*(3y^2+18y-48)/(y^2-11y+18) The quadratic expressions in the above rational factorise as, {(2*(3*y+1))*(y-9) / (y+8)*(3*y+1)} * {(3*(y+8))*(y-2) / (y-2)*(y-9)}   which simplifies to {(2*(y-9) / (y+8)} * {(3*(y+8)) / (y-9)}  looking closely at the above expression, we can see that all bracketed terms cancel, leaving us with {2} * {3} The rational expression reduces down to: 6

### 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.

### how do you use brackets in simultaneous equations

One reason you might use brackets in simultaneous equations is for expressing the solution as an ordered set (usually pairs). If, for example, you had a 2-variable system of equations, x and y, you might express the result as the ordered pair (x,y) where x is replaced by the value found for x and y the value found for y. Sometimes a system of equations has more than one solution, so the different solutions would be represented as (x1,y1) and (x2,y2). This can happen when the system involves one or more quadratic equations. The brackets ensure that the values for x and y are not mixed up. Another reason for brackets is when using substitution to solve a system. Suppose there are two equations: ax+by=c and dx+ey=f. From the first equations we can write y=(c-ax)/b and substitute for y in the second equation: dx+e(c-ax)/b=f. That would be the first step in solving. The next step would be to expand the brackets and solve for x in terms of the constants a, b, c, d, e and f. Having found x, you find y by substituting the value of x in y=(c-ax)/b. Sometimes a question involving simultaneous equations comes in a form that uses brackets, so the first step is to expand the brackets, combine like terms, and then continue to solve the system.

### Find the point(s) where the line through the origin with slope 6 intersects the unit circle.

The line passes through the origin so its equation is y=6x. The equation of the unit circle is x^2+y^2=1, which has centre (0,0) and radius 1. Substituting y=6x in the equation of the circle we have 37x^2=1 and x=+sqrt(1/37). Therefore the y values for the intersections are y=+6sqrt(1/37). The points of intersection are (-sqrt(1/37),-6sqrt(1/37)), (sqrt(1/37),6sqrt(1/37)).  If the unit circle has centre (h,k) the equation is (x-h)^2+(y-k)^2=1 and substituting y=6x we get (x-h)^2+(6x-k)^2=1, which gives the x value of the intersection. So x^2-2xh+h^2+36x^2-12xk+k^2=1; 37x^2-2x(h+6k)+h^2+k^2-1=0. There are only two factors of 37, which is prime, so to factorise rationally we must have (37x+a)(x+b)=0; 37x^2+x(37b+a)+ab=0. Using the quadratic formula: x=(h+6k+sqrt((h+6k)^2-37(h^2+k^2-1))/37. The square root can only be evaluated if the expression is positive, so (h+6k)^2>37(h^2+k^2-1). This requirement applies so that the line intersects the unit circle. When the expression is zero, the line is a tangent to the circle, so there is only one intersection point. h^2+36k^2+12hk>37h^2+37k^2-37; 36h^2-12hk+k^2<37; (6h-k)^2<37 and (6h-k)< +sqrt(37) (=+6.08). This connects the coordinates of the centre of the unit circle: k>6h+sqrt(37). On equality the line y=6x will be tangential to the circle. For example, if h=0 (centre of the circle is on the y axis), k=+sqrt(37) and the unit circle will lie above or below the axis with y=6x running tangentially on the right of the circle; or on the left touching the circle below the x axis. Ideally, we want the square root to be rational so 37-(6h-k)^2=a^2. If a=+1, 6h-k=6 and k=6(h-1); or 6h-k=-6, so k=6(h+1). That gives many possible values for h and k represented by pairs: (1,0), (2,6), (3,12), (0,-6), (-1,-12), (-2,-18), (0,6), (1,12), (-1,0), (-2,-6),  to mention but a few. Using (1,0) in the quadratic: 37x^2-2x(h+6k)+h^2+k^2-1=0 we have 37x^2-2x=0=x(37x-2) giving intersection points (0,0) and (2/37,12/37). The equation for the circle is (x-1)^2+y^2=1 or y^2=2x-x^2. Let's try (2,6). 37x^2-76x+39=0, (37x-39)(x-1)=0 giving intersection points (39/37,234/37) and (1,6). The equation of the circle is (x-2)^2+(y-6)^2=1 or x^2-4x+y^2-12x+39=0. If a=+6, 6h-k=1 or -1, so k=6h-1 or 6h+1. This generates more possible intersection points. There are clearly an infinite number of positions for the unit circle centre (h,k) and an infinite number of intersection points. However, the relationship between h and k so as to produce rational intersection points has been established. k=6(h+1), k=6h+1 are the equations linking the coordinates of the centre of the unit circle. With these equations in mind the quadratic determining the intersection points (x,6x) can be solved: 37x^2-2x(h+6k)+h^2+k^2-1=0. There are four variations of this quadratic because there are four equations linking h and k. Recap There are 2 values of a^2 where a^2=37-(6h-k)^2 and x=(h+6k+a)/37; a^2=1 or 36. When a^2=1, k=6(h+1). The equation of the circle is (x-h)^2+(y-6(h+1))^2=1 and x=(37h+36+1)/37. So the points of intersection are (h+1,6(h+1)), ((37h+35)/37,6(37h+35)/37), ((37h-35)/37,6(37h-35)/37), (h-1,6(h-1)). When a^2=36, k=6h+1. The equation of the circle is (x-h)^2+(y-6h+1)^2=1 and x=(37h+6+6)/37. The points of intersection are ((37h+12)/37,6(37h+12)/37), (h,6h), ((37h-12)/37,6(37h-12)/37). Note that (h,6h) is the result of (37h+6-6)/37 and (37h-6+6)/37. We can check the (h,k) values we used earlier. These were (1,0) and (2,6). We used the formula k=6(h-1) in each case (a=+1), so intersection points for h=1, k=6(h-1)=0, should be x=(h+6k+1)/37, giving (2/37,12/37) and x=(h+6k-1), giving (0,0). For (2,6) h=2 and k=6, giving intersection points x=(2+36+1)/37, giving (39/37,234/37) and x=(2+36-1)/37=1, giving (1,6). The values of h and k are not restricted to integers.

### Do sequences with an equivalent second difference have a quadratic nth term?

Do sequences with an equivalent second difference have a quadratic nth term? Yes they do. When the 1st differences are constant, then we have an arithmetic sequence, with the nth term given by a linear function. When it is the 2nd differences that are constsant, then the nth term is given by a quadratic function. See below.   Compound Arithmetic Sequence         b0          b1         b2          b3          b4           b5                c0          c1          c2          c3          c4  -------------- 1st differences                        d            d            d             d   ------------------- 2nd differences = constant We have here an irregular sequence (b_n) = (b_1, b_2, b_3, ..., b_k, ...). The (1st) differences between the elements of (b_n), the 1st differences, are non-constant. However, the (2nd) differences between the elements of (c_n) are constant. This makes (c_n) a regular arithmetic sequence, and so we can write, cn = c0 + nd,  n = 0,1,2,3,... The sequence (b_n) is non-regular, but we can write, b_(n+1) = b_n + c_n Using the expression for the c-sequence, b_(n+1) = b_n + c_0 + nd Solve the recurrence relation for b_n Develop the terms of the sequence. b1 = b0 + c0 b2 = b1 + c0 + 1.d = b0 + 2.c0 + 1.d b3 = b2 + c0 + 2.d = b0 + 3.c0 + (1 + 2).d b4 = b3 + c0 + 3.d = b0 + 4.c0 + (1 + 2 + 3).d ‘ ‘ ‘ b(n+1) = b_0 + (n+1).c0 + sum[k=1..n](k) * d b_(n+1) = b_0 + (n+1).c0 + (nd/2)(n + 1) b_(n+1) = b_0 + (n+1)(c_0 + nd/2) We could also write c0 = b1 – b0 in the above expression.

Let f(x)=36x^2-12x+a^2-b^2=36x^2-12x+1+a^2-b^2-1=(6x-1)^2+(a-b)(a+b)-1. If a^2-b^2=1, the quadratic expression f(x)=(6x-1)^2. When f(x)=0, x=1/6. The general solution is (6x-1)^2=1+b^2-a^2, so 6x-1=+sqrt(1+b^2-a^2), making x=(1/6)(1+sqrt(1+b^2-a^2)). If b=any number, and a=+1, the square root disappears and x=(1/6)(1+b). Or just considering factorisation: f(x)=(6x-1)^2-b^2=(6x-1-b)(6x-1+b), and a=1.

### use a graphical method to solve this simultaneous equation: y=x^2 and y=x+6

Plot y=x^2 and y=x+6 on the same graph. The first is a U-shaped curve sitting on the origin (0,0), the second is a straight line crossing the x axis (y=0) at x=-6 and crossing the y axis (x=0) at y=6. Where the line cuts through the U curve represents the solution to the equation x^2-x-6=0 (x^2=x+6). The solution to this quadratic is x=-2, +3. The intersection points are therefore (-2, 4) and (3, 9). The remaing part of your question involves straight lines only, linear equations. For (a) draw a straight line joining y=-4 on the y axis to x=2 on the x axis. Also draw a line joining y=3 1/3 to x=-1 2/3. These points on the axes are where x=0 and y=0 for the two functions. The lines don't stop at the axes, so just continue them after they cut the axes. What may have been confusing to you is that the first equation starts y=..., but the second equation has x and y together in an expression. However, you can move things around in an equation and, if you want, you can make the equation look like y=... or x=... or just combine x and y in an expression. It doesn't matter. What you'll find is that the lines are parallel (have the same slope) so that means they never cross and that means there's no solution. In (b) the two lines have different slopes so we would expect them to intersect. The first equation means joining (0,3) to (9/2,0) and (0,11/2) to (11/3,0). Again, extend the lines to beyond where they cut the axes. Note that the second equation simplifies to 3x+2y=11. The solution to the equations solved simultaneously or by substitution give a single intersection point at (3,1).

### Show me how to solve the diff eqn m^4-2αm^2-β=0

Show me how to solve the diff eqn m^4-2αm^2-β=0 This is not a differential equation. It is a quadratic expression in m^2. Let x = m^2, then the equation is x^2 – 2αx – β = 0 Using the quadratic formula, x = (2α ± √((-2α)^2 – 4.1. (-β)) / (2*1) x = (2α ± √(4α^2 + 4β) / (2) x = α ± √(α^2 + β) x = α + √(α^2 + β), X = α - √(α^2 + β) But x = m^2, therefore m = ± √( α + √(α^2 + β)), m = ± √( α - √(α^2 + β))

### Solve -x^3+2x^2+9x+22

Solve –x^3 + 2x^2 + 9x + 22 I need to find the real and complex solutions. This is a cubic equation, with real and complex solutions. There will be two complex solutions and one real solution. The real solution Use the Newton-Raphson method on f(x)= -x^3+2x^2+9x+22 The N-R formula to use is: x_(n+1)=x_n - f(x_n )/(f^' (x_n ) ) Where f(x)= -x^3+2x^2+9x+22,  f^' (x)=-3x^2+4x+9, and x_1=4. (found by trial and error) n |     x_n    |       f(x_n)        |  f^' (x_n ) | x_(n+1)=x_n-f(x_n )/(f^' (x_n ) ) 1 |   4          |      26              |   -23        | 5.13043 2 | 5.13043 | -14.2234         | -49.4423 | 4.84276 3 | 4.84276 | -1.08443         | -41.9859 | 4.81693 4 | 4.81693 | -8.3404 E -03 | -41.3407 | 4.81673 5 | 4.81673 | -5.0677 E -07 | -41.3357 | 4.81673 So, to 5 d.p. our real solution is: x=4.81673 Our original expression is: -x^3+2x^2+9x+22. Taking out a factor of (x-4.81673), we get (x - 4.81673)(-x^2 - 2.81673x - 4.567416) The remaining quadratic is: g(x)=-x^2 - 2.81673x - 4.567416. Solving this with the quadratic formula, x=(-b±√(b^2-4ac))/2a With a=-1, b=-2.81673, c=-4.567416. Then, x=(2.81673±√(2.81673^2 - 4(-1)(-4.567416) ))/(2*(-1) ) x=(2.81673±√(7.93396 - 18.26966))/(-2) x=(2.81673±i√10.33571)/(-2) x = -1.40836 ± 1.60746i