What are all the names for solutions to quadratic equations?
Q. I'm doing a geometry review for my students that I am subbing, and I'm supposed to know the names for solutions to quadratic equations, any ideas? Please and thank you!
Asked by Sarah - Tue Nov 4 21:22:12 2008 - - 1 Answers - 0 Comments
A. Do you mean like roots, zeroes, solutions?
Answered by Matt T - Tue Nov 4 21:24:54 2008
Q. I'm doing a geometry review for my students that I am subbing, and I'm supposed to know the names for solutions to quadratic equations, any ideas? Please and thank you!
Asked by Sarah - Tue Nov 4 21:22:12 2008 - - 1 Answers - 0 Comments
A. Do you mean like roots, zeroes, solutions?
Answered by Matt T - Tue Nov 4 21:24:54 2008
When are quadratic equations used in real life?
Q. I have a project where I have to write a 5 page essay about quadratic equations. I have heard about people using them for missiles in the army, and satellite dishes. If you have any information, please, PLEASE, help me.
Asked by HELP ME - Sun Dec 14 18:44:36 2008 - - 3 Answers - 0 Comments
A. Yes, missiles and other projectiles under the influence of gravity (or some other accelerating agent), will follow a parabolic trajectory (if we ignore air resistance and some other factors). For example, at time t, the vertical height of an object launched into the air will be: x(t) = (1/2)*g*t^2 + v0*t + x0 where t represents the time since the object was launched, g represents the acceleration due to gravity, v0 represents the objects initial vertical velocity, and x0 represents the objects initial vertical height. For more information, try searching for "projectile motion," "free-fall motion," "trajectory," etc.
Answered by greentunic - Sun Dec 14 19:03:14 2008
Q. I have a project where I have to write a 5 page essay about quadratic equations. I have heard about people using them for missiles in the army, and satellite dishes. If you have any information, please, PLEASE, help me.
Asked by HELP ME - Sun Dec 14 18:44:36 2008 - - 3 Answers - 0 Comments
A. Yes, missiles and other projectiles under the influence of gravity (or some other accelerating agent), will follow a parabolic trajectory (if we ignore air resistance and some other factors). For example, at time t, the vertical height of an object launched into the air will be: x(t) = (1/2)*g*t^2 + v0*t + x0 where t represents the time since the object was launched, g represents the acceleration due to gravity, v0 represents the objects initial vertical velocity, and x0 represents the objects initial vertical height. For more information, try searching for "projectile motion," "free-fall motion," "trajectory," etc.
Answered by greentunic - Sun Dec 14 19:03:14 2008
Why are there usually two solutions in quadratic equations?
Q. Hi need help finding the answer to this question Please? Why are there usually two solutions in quadratic equations?
Asked by chianti72116 - Wed Jul 22 17:26:47 2009 - - 3 Answers - 0 Comments
A. The quadratic equation is as follows: A * x^2 + B * x + C Where A, B, and C are coefficients. The quadratic formula (solving for x) is this: (-B +/- sqrt(B^2 - 4*A*C)) / (2A) This means that x = (-B + sqrt(B^2 - 4*A*C)) / (2A) and x = (-B - sqrt(B^2 - 4*A*C)) / (2A) Thus you have 2 solutions (and in any polynomial of the "N"th degree, you can have at most an N amount of real and/or imaginary solutions, but you do not necessarily have to have an N amount of solutions). However, if B^2 - 4*A*C < 0, then you will have 2 imaginary solutions and no real solutions. If B^2 - 4*A*C >/= 0, then you will have 2 real solutions. It's one set or the other, with nothing in between (i.e. one real and one fake solution, this does not ever… [cont.]
Answered by Captain Matticus, LandPiratesInc - Wed Jul 22 17:38:13 2009
Q. Hi need help finding the answer to this question Please? Why are there usually two solutions in quadratic equations?
Asked by chianti72116 - Wed Jul 22 17:26:47 2009 - - 3 Answers - 0 Comments
A. The quadratic equation is as follows: A * x^2 + B * x + C Where A, B, and C are coefficients. The quadratic formula (solving for x) is this: (-B +/- sqrt(B^2 - 4*A*C)) / (2A) This means that x = (-B + sqrt(B^2 - 4*A*C)) / (2A) and x = (-B - sqrt(B^2 - 4*A*C)) / (2A) Thus you have 2 solutions (and in any polynomial of the "N"th degree, you can have at most an N amount of real and/or imaginary solutions, but you do not necessarily have to have an N amount of solutions). However, if B^2 - 4*A*C < 0, then you will have 2 imaginary solutions and no real solutions. If B^2 - 4*A*C >/= 0, then you will have 2 real solutions. It's one set or the other, with nothing in between (i.e. one real and one fake solution, this does not ever… [cont.]
Answered by Captain Matticus, LandPiratesInc - Wed Jul 22 17:38:13 2009
How do you solve the nth term for Quadratic Equations?
Q. I really need to know this. I am in top set for maths so I know how to do linear equations and that, but when I look on yahoo for this question, the answer aren't explained very clearly. This one answer started multiplying by 4 and they never said why, then they subtracted and I couldn't understand it. Please explain in detail what you need to do in order to find the nth term for a quadratic sequence. Explain it simply too!
Asked by The Wrestling World - Wed Apr 29 16:03:23 2009 - - 1 Answers - 0 Comments
A. Here's an example sequence : 5, 12, 23, 38, 57, ... Take the 1st differences : 12 - 5 = 7 23 - 12 = 11 38 - 23 = 15 57 - 38 = 19 Line up the answers : 7, 11, 15, 19, ... Now take the 2nd differences : 11 - 7 = 4 15 - 11 = 4 19 - 15 = 4 All the 2nd differences are the same, so this will be a quadratic expression. (If instead, the 1st differences had been the same, then a linear expression would do) (If the 3rd differences had been the same, then a cubic expression would be needed) (etc.) So we try to fit the data to the quadratic expression, which is the n(th) term : an^2 + bn + c Now we have to find the values of a, b and c. When n = 1, we want an^2 + bn + c to equal 5. When n = 2, we want an^2 + bn + c to equal 12. etc. So,… [cont.]
Answered by falzoon - Thu Apr 30 10:09:55 2009
Q. I really need to know this. I am in top set for maths so I know how to do linear equations and that, but when I look on yahoo for this question, the answer aren't explained very clearly. This one answer started multiplying by 4 and they never said why, then they subtracted and I couldn't understand it. Please explain in detail what you need to do in order to find the nth term for a quadratic sequence. Explain it simply too!
Asked by The Wrestling World - Wed Apr 29 16:03:23 2009 - - 1 Answers - 0 Comments
A. Here's an example sequence : 5, 12, 23, 38, 57, ... Take the 1st differences : 12 - 5 = 7 23 - 12 = 11 38 - 23 = 15 57 - 38 = 19 Line up the answers : 7, 11, 15, 19, ... Now take the 2nd differences : 11 - 7 = 4 15 - 11 = 4 19 - 15 = 4 All the 2nd differences are the same, so this will be a quadratic expression. (If instead, the 1st differences had been the same, then a linear expression would do) (If the 3rd differences had been the same, then a cubic expression would be needed) (etc.) So we try to fit the data to the quadratic expression, which is the n(th) term : an^2 + bn + c Now we have to find the values of a, b and c. When n = 1, we want an^2 + bn + c to equal 5. When n = 2, we want an^2 + bn + c to equal 12. etc. So,… [cont.]
Answered by falzoon - Thu Apr 30 10:09:55 2009
How would I figure out where to plot points when doing quadratic equations, with out using my calculator?
Q. Is their a way to figure out all points on a graph when doing quadratic equations such as: y=x^2+4x+6,y=2x^2+4,y=3x^ 2,y=x^2-9,or y=(x-4)^2. When I do them on my calculator I can't tell all points. Again my calculator is a Casio fx-9750 ga plus power Graphic.
Asked by Cheryl - Tue Aug 25 00:27:37 2009 - - 2 Answers - 0 Comments
A. y=x^2+4x+6 = x^2+4x+4+2 =(x+2)^2+2 For x=-2, y=2 x=-1, y=3 x=0, y= 6 x=1, y= 11 x=2 y= 18 y=2x^2+4 For x=-2, y= 12 x=-1, y= 6, x=0, y= 4 x=1, y= 6 x=2, y= 12 y=3x^2 For x=-3, y= 27 x=-2, y= 12 x=-1, y= 3 x=0, y=0 x=1, y= 3 x=2, y= 12 x=3, y=27 y=x^2-9=(x+3)(x-3) For x=-3, y=0 x=-2, y=-5 x=-1, y= -8 x=0, y= -9 x=1, y= -8,x=2, y= -5 x=2, y=-5 x=3, y=0 y=(x-4)^2. For x=4, y=0 x=3, y= 1 x=2, y= 4 x=1, y= 9 x=0, y= 16 x=-1, y= 25 x=-2, y=36
Answered by DOVE - Tue Aug 25 00:40:04 2009
Q. Is their a way to figure out all points on a graph when doing quadratic equations such as: y=x^2+4x+6,y=2x^2+4,y=3x^ 2,y=x^2-9,or y=(x-4)^2. When I do them on my calculator I can't tell all points. Again my calculator is a Casio fx-9750 ga plus power Graphic.
Asked by Cheryl - Tue Aug 25 00:27:37 2009 - - 2 Answers - 0 Comments
A. y=x^2+4x+6 = x^2+4x+4+2 =(x+2)^2+2 For x=-2, y=2 x=-1, y=3 x=0, y= 6 x=1, y= 11 x=2 y= 18 y=2x^2+4 For x=-2, y= 12 x=-1, y= 6, x=0, y= 4 x=1, y= 6 x=2, y= 12 y=3x^2 For x=-3, y= 27 x=-2, y= 12 x=-1, y= 3 x=0, y=0 x=1, y= 3 x=2, y= 12 x=3, y=27 y=x^2-9=(x+3)(x-3) For x=-3, y=0 x=-2, y=-5 x=-1, y= -8 x=0, y= -9 x=1, y= -8,x=2, y= -5 x=2, y=-5 x=3, y=0 y=(x-4)^2. For x=4, y=0 x=3, y= 1 x=2, y= 4 x=1, y= 9 x=0, y= 16 x=-1, y= 25 x=-2, y=36
Answered by DOVE - Tue Aug 25 00:40:04 2009
What is the ratio of the quadratic equations with real solutions to those having non-real solutions?
Q. or for every 100 quadratic equations how many have real solutions.
Asked by fauzdar65 - Fri Oct 24 04:46:05 2008 - - 2 Answers - 0 Comments
A. There is an infinite number of quadratic equations with real solutions There is also an infinite number of quadratic equations with non-real solutions Because you asked that question, I would not try to explain the relative order of infinities for both cases.
Answered by Jerome J - Fri Oct 24 05:17:31 2008
Q. or for every 100 quadratic equations how many have real solutions.
Asked by fauzdar65 - Fri Oct 24 04:46:05 2008 - - 2 Answers - 0 Comments
A. There is an infinite number of quadratic equations with real solutions There is also an infinite number of quadratic equations with non-real solutions Because you asked that question, I would not try to explain the relative order of infinities for both cases.
Answered by Jerome J - Fri Oct 24 05:17:31 2008
Are there quadratic equations in calculus?
Q. Are there quadratic equations in calculus? Why are they called quadratic equations, what does quad-rat-ic denote ? It is understood to mean what?
Asked by Snow Bird - Mon Jul 27 01:00:47 2009 - - 5 Answers - 0 Comments
A. most of calculus deal with quadratic equations ( quadratic means squares or 2nd power)
Answered by Jason W - Mon Jul 27 01:10:32 2009
Q. Are there quadratic equations in calculus? Why are they called quadratic equations, what does quad-rat-ic denote ? It is understood to mean what?
Asked by Snow Bird - Mon Jul 27 01:00:47 2009 - - 5 Answers - 0 Comments
A. most of calculus deal with quadratic equations ( quadratic means squares or 2nd power)
Answered by Jason W - Mon Jul 27 01:10:32 2009
Has anybody used quadratic equations in everyday situations?
Q. Talking to my friends about subjects we were taught in school nobody has used quadratic equations or could think of a situation where they would be of any use.
Asked by john.brookes14@btinternet.com - Mon Oct 1 14:23:17 2007 - - 2 Answers - 1 Comments
A. While the specifics (use of the quadratic equation, etc) may have fallen into disuse, the process of categorizing a problem, determining the most appropriate method to solve it, applying this method, and checking the solution which came hand-in-hand with learning problem-solving in mathematics is a process you likely use every day.
Answered by BNP - Mon Oct 1 14:33:17 2007
Q. Talking to my friends about subjects we were taught in school nobody has used quadratic equations or could think of a situation where they would be of any use.
Asked by john.brookes14@btinternet.com - Mon Oct 1 14:23:17 2007 - - 2 Answers - 1 Comments
A. While the specifics (use of the quadratic equation, etc) may have fallen into disuse, the process of categorizing a problem, determining the most appropriate method to solve it, applying this method, and checking the solution which came hand-in-hand with learning problem-solving in mathematics is a process you likely use every day.
Answered by BNP - Mon Oct 1 14:33:17 2007
Anyone good with parabolas and quadratic equations?
Q. A parabola is tangent to the x-axis, has a y-intercept of 18 and passes through the point (2,8). There are two possible quadratic equations, find both. Can someone help me with my math IQ's?
Asked by Justine - Mon Dec 15 19:52:19 2008 - - 1 Answers - 0 Comments
A. "There are two possible quadratic equations, find both"? Hum, I've never heard of that. I guess I've always assumed there is only one that fits these parameters. Anyway, I'll show you how to find the obvious one and maybe you can figure out the other (if there is another). But ask your teacher about this for sure. So you have two points given and a third assumed, and from these three points we will create the quadratic function. The first point of course is (2,8). The second is the y-intercept = 18 and so that is also obviously (0,18). The third is assumed as (8,2), which makes this parabola tangent to the x-axis (meaning it touches it in only one place). This is also expressed by: b - 4ac > 0 So OK, we have our three points and we… [cont.]
Answered by Don - Mon Dec 15 21:05:25 2008
Q. A parabola is tangent to the x-axis, has a y-intercept of 18 and passes through the point (2,8). There are two possible quadratic equations, find both. Can someone help me with my math IQ's?
Asked by Justine - Mon Dec 15 19:52:19 2008 - - 1 Answers - 0 Comments
A. "There are two possible quadratic equations, find both"? Hum, I've never heard of that. I guess I've always assumed there is only one that fits these parameters. Anyway, I'll show you how to find the obvious one and maybe you can figure out the other (if there is another). But ask your teacher about this for sure. So you have two points given and a third assumed, and from these three points we will create the quadratic function. The first point of course is (2,8). The second is the y-intercept = 18 and so that is also obviously (0,18). The third is assumed as (8,2), which makes this parabola tangent to the x-axis (meaning it touches it in only one place). This is also expressed by: b - 4ac > 0 So OK, we have our three points and we… [cont.]
Answered by Don - Mon Dec 15 21:05:25 2008
Why aren't Quadratic equations calculated using diameter of a circle that grows exponentially ?
Q. I was just curious why they don't have a formula for calculating a quadratic equation using circle measurements. You would think that there is a certain rate of growth that would occur exponentially. Like stacking circles over another circle.It seems like this would be just as much a geometric function than an algebraic function. Just confused, anyways thanks.
Asked by Isaac O - Wed Jan 28 09:09:17 2009 - - 2 Answers - 0 Comments
Q. I was just curious why they don't have a formula for calculating a quadratic equation using circle measurements. You would think that there is a certain rate of growth that would occur exponentially. Like stacking circles over another circle.It seems like this would be just as much a geometric function than an algebraic function. Just confused, anyways thanks.
Asked by Isaac O - Wed Jan 28 09:09:17 2009 - - 2 Answers - 0 Comments
How do I find quadratic equations from roots?
Q. I am given the roots 1+ - i (One plus or Minus i) and I need to find the quadratic equation that goes to it. Then later on I need to find a cubic equation with itegral coefficients that have the given roots... i sqrt2 and 5 If you have any idea please help.
Asked by Zeriath C - Mon Oct 20 01:12:01 2008 - - 1 Answers - 0 Comments
A. ok if a and b are 2 roots of a certain equation then (x - a)(x - b) = 0 apply for a = 1 + i and b = 1 - i, expand the left side to get the original equation whose roots are a and b, remember i^2 = -1 same for the other
Answered by 'est la vie - Mon Oct 20 01:19:13 2008
Q. I am given the roots 1+ - i (One plus or Minus i) and I need to find the quadratic equation that goes to it. Then later on I need to find a cubic equation with itegral coefficients that have the given roots... i sqrt2 and 5 If you have any idea please help.
Asked by Zeriath C - Mon Oct 20 01:12:01 2008 - - 1 Answers - 0 Comments
A. ok if a and b are 2 roots of a certain equation then (x - a)(x - b) = 0 apply for a = 1 + i and b = 1 - i, expand the left side to get the original equation whose roots are a and b, remember i^2 = -1 same for the other
Answered by 'est la vie - Mon Oct 20 01:19:13 2008
How to factorize a quadratic equations?
Q. I need a complete step-by-step on how to factorize a quadratic equations because i'm totally clueless about it.And please make it very simple to for me understand.Thanks...
Asked by xsplinter_89 - Fri Mar 9 03:26:11 2007 - - 5 Answers - 0 Comments
A. Try these, I hope they help you --- Let's look at an example: x + 6x + 8 = 0. We can rewrite it as x + 2x + 4x + 8 = 0. We can factorise this into two terms: x (x + 2) + 4 (x + 2) = 0. Notice that (x + 2) appears twice: it is a common factor! We can factorise the expression further to: (x + 4)(x + 2) = 0 giving the correct solutions: x=-4 and x=-2. You may have spotted that 4 + 2 = 6 and that 4 2 = 8. Similar quadratic equations can be factorised and solved if the number on its own (c in the standard equation) has factors that add up to b (the number of x's). --- Solving quadratic equations by factorising. If we have two numbers, A and B, such that AxB = 0, then it must follow that either A = 0, or B = 0 (or both), because… [cont.]
Answered by NineLivesBurra - Fri Mar 9 03:37:31 2007
Q. I need a complete step-by-step on how to factorize a quadratic equations because i'm totally clueless about it.And please make it very simple to for me understand.Thanks...
Asked by xsplinter_89 - Fri Mar 9 03:26:11 2007 - - 5 Answers - 0 Comments
A. Try these, I hope they help you --- Let's look at an example: x + 6x + 8 = 0. We can rewrite it as x + 2x + 4x + 8 = 0. We can factorise this into two terms: x (x + 2) + 4 (x + 2) = 0. Notice that (x + 2) appears twice: it is a common factor! We can factorise the expression further to: (x + 4)(x + 2) = 0 giving the correct solutions: x=-4 and x=-2. You may have spotted that 4 + 2 = 6 and that 4 2 = 8. Similar quadratic equations can be factorised and solved if the number on its own (c in the standard equation) has factors that add up to b (the number of x's). --- Solving quadratic equations by factorising. If we have two numbers, A and B, such that AxB = 0, then it must follow that either A = 0, or B = 0 (or both), because… [cont.]
Answered by NineLivesBurra - Fri Mar 9 03:37:31 2007
Can you help me with these Quadratic Equations please?
Q. Hi, I need help on Quadratic Equations please. Factorise: 1) x squared - 6x+8 Solve the Equation 2) x squared - 6x+8=0 The squared is meant to be a little two next to the x but you know I can't do that. Thank You!
Asked by Jerry - Tue Mar 4 16:59:29 2008 - - 5 Answers - 0 Comments
A. x^2 - 6x + 8 = 0 (x - 4)(x - 2) = 0. x = 4. x = 2. Hope this helps.
Answered by John - Tue Mar 4 17:05:47 2008
Q. Hi, I need help on Quadratic Equations please. Factorise: 1) x squared - 6x+8 Solve the Equation 2) x squared - 6x+8=0 The squared is meant to be a little two next to the x but you know I can't do that. Thank You!
Asked by Jerry - Tue Mar 4 16:59:29 2008 - - 5 Answers - 0 Comments
A. x^2 - 6x + 8 = 0 (x - 4)(x - 2) = 0. x = 4. x = 2. Hope this helps.
Answered by John - Tue Mar 4 17:05:47 2008
Having trouble trying to answer these two math questions about quadratic equations?
Q. I am having trouble trying to answer these two math questions? a) What can we tell about the graph of a quadratic equation just by examining the equation (with minor calculations)? How can we tell these things? b) What are some applications of quadratic equations? If these were actual quadratic equation problems it would be easier to do but I have to type out at least 3 sentences. Can anyone provide me with some websites to research?? Thanks
Asked by cuvelx - Sun Dec 16 00:51:14 2007 - - 1 Answers - 0 Comments
A. "If a, b, and c are real numbers, and the domain of f is the set of real numbers, then the zeros of f are exactly the x-coordinates of the points where the graph touches the x-axis. It follows from the above that, if the discriminant is positive, the graph touches the x-axis at two points, if zero, the graph touches at one point, and if negative, the graph does not touch the x-axis."
Answered by Rafaelinux! - Sun Dec 16 00:56:58 2007
Q. I am having trouble trying to answer these two math questions? a) What can we tell about the graph of a quadratic equation just by examining the equation (with minor calculations)? How can we tell these things? b) What are some applications of quadratic equations? If these were actual quadratic equation problems it would be easier to do but I have to type out at least 3 sentences. Can anyone provide me with some websites to research?? Thanks
Asked by cuvelx - Sun Dec 16 00:51:14 2007 - - 1 Answers - 0 Comments
A. "If a, b, and c are real numbers, and the domain of f is the set of real numbers, then the zeros of f are exactly the x-coordinates of the points where the graph touches the x-axis. It follows from the above that, if the discriminant is positive, the graph touches the x-axis at two points, if zero, the graph touches at one point, and if negative, the graph does not touch the x-axis."
Answered by Rafaelinux! - Sun Dec 16 00:56:58 2007
Can you solve these quadratic equations by factoring?
Q. Can you please solve these quadratic equations? (k+1)(K-5)=0 (A+1)(A+2=0 (4k+5)(k+1)=0 (2m+3)(4m+3)=0 x2-11x+19=-5 n+7n+15=5 n2-10n+22=-2 n2+3n-12=6 6n2-18n-18=6 7r2-14r=-7 please explain.
Asked by !My*HeArT*Iz*DaMaGeD! - Sun May 10 12:42:33 2009 - - 3 Answers - 0 Comments
A. Im in 7th grade and i just learnt it. Use the foil method. First,k times K, Outer, K times negative 5, Inner, 1 times K, Last, 1 times negative 5. Hope it helps
Answered by [PROTOYPE] - Sun May 10 12:55:18 2009
Q. Can you please solve these quadratic equations? (k+1)(K-5)=0 (A+1)(A+2=0 (4k+5)(k+1)=0 (2m+3)(4m+3)=0 x2-11x+19=-5 n+7n+15=5 n2-10n+22=-2 n2+3n-12=6 6n2-18n-18=6 7r2-14r=-7 please explain.
Asked by !My*HeArT*Iz*DaMaGeD! - Sun May 10 12:42:33 2009 - - 3 Answers - 0 Comments
A. Im in 7th grade and i just learnt it. Use the foil method. First,k times K, Outer, K times negative 5, Inner, 1 times K, Last, 1 times negative 5. Hope it helps
Answered by [PROTOYPE] - Sun May 10 12:55:18 2009
How do you solve these quadratic equations?
Q. Solve the following quadratic equations. (And try to include your steps and explain) (x-2)^2 - 2(x-2) - 15 = 0 (2x-3)^2 - 3(2x-3) -10 = 0 Any help is appreciated! :)
Asked by Audge Podge. - Sat Mar 28 10:50:32 2009 - - 6 Answers - 0 Comments
A. GIVEN (x-2)^2 - 2(x-2) - 15 = 0 Step 1 -- expand the above x^2 - 4x + 4 - 2x + 4 - 15 = 0 Step 2 -- perform the required arithmetic x^2 - 6x - 7 = 0 Step 3 -- factor the above expression (x - 7)(x + 1) = 0 Step 4 -- solve for the roots x = 7 and x = -1 For your second problem, simply follow the above steps. Hope this helps.
Answered by Pointy - Sat Mar 28 10:55:56 2009
Q. Solve the following quadratic equations. (And try to include your steps and explain) (x-2)^2 - 2(x-2) - 15 = 0 (2x-3)^2 - 3(2x-3) -10 = 0 Any help is appreciated! :)
Asked by Audge Podge. - Sat Mar 28 10:50:32 2009 - - 6 Answers - 0 Comments
A. GIVEN (x-2)^2 - 2(x-2) - 15 = 0 Step 1 -- expand the above x^2 - 4x + 4 - 2x + 4 - 15 = 0 Step 2 -- perform the required arithmetic x^2 - 6x - 7 = 0 Step 3 -- factor the above expression (x - 7)(x + 1) = 0 Step 4 -- solve for the roots x = 7 and x = -1 For your second problem, simply follow the above steps. Hope this helps.
Answered by Pointy - Sat Mar 28 10:55:56 2009
How do I write a quadratic equations with the roots having "i"?
Q. Im trying to write a quadratic equation in standard form: ax^2+bx+c=0 with the answer being : -5 + or - 8i so i remembered this might work: [x-(-5+8i)] [x-(-5-8i)] but i don't know how to solve this properly. It would be very nice if someone could help show me step by step. Thank you =D
Asked by zanfankid - Wed Nov 19 02:18:39 2008 - - 3 Answers - 0 Comments
A. You are correct that it will be [x - (-5 + 8i)]*[x - (-5 - 8i)] = 0 Now multiply out to get x^2 - (-5 + 8i)x - (-5 - 8i)x + (-5 + 8i)(-5 - 8i) = 0 Can you finish it from there? All the "i"s disappear.
Answered by mathsmanretired - Wed Nov 19 02:33:30 2008
Q. Im trying to write a quadratic equation in standard form: ax^2+bx+c=0 with the answer being : -5 + or - 8i so i remembered this might work: [x-(-5+8i)] [x-(-5-8i)] but i don't know how to solve this properly. It would be very nice if someone could help show me step by step. Thank you =D
Asked by zanfankid - Wed Nov 19 02:18:39 2008 - - 3 Answers - 0 Comments
A. You are correct that it will be [x - (-5 + 8i)]*[x - (-5 - 8i)] = 0 Now multiply out to get x^2 - (-5 + 8i)x - (-5 - 8i)x + (-5 + 8i)(-5 - 8i) = 0 Can you finish it from there? All the "i"s disappear.
Answered by mathsmanretired - Wed Nov 19 02:33:30 2008
why do we set quadratic equations equal to zero?
Q. how come when we see quadratic equations we have to set it equal to zero before we start solving.
Asked by dk d - Wed Jan 2 14:49:45 2008 - - 5 Answers - 0 Comments
A. This page explains it well -- enjoy
Answered by free_your_fancy - Wed Jan 2 14:56:25 2008
Q. how come when we see quadratic equations we have to set it equal to zero before we start solving.
Asked by dk d - Wed Jan 2 14:49:45 2008 - - 5 Answers - 0 Comments
A. This page explains it well -- enjoy
Answered by free_your_fancy - Wed Jan 2 14:56:25 2008
How do you solve quadratic equations?
Q. I need to know how to solve this quadratic equation: 12x^2-15x= 0 (The directions on my math packet say if the equation is not factor-able, use the quadratic formula to find the roots. What does this mean?)
Asked by Hayy-layyy :] - Mon Nov 3 00:40:17 2008 - - 6 Answers - 0 Comments
Q. I need to know how to solve this quadratic equation: 12x^2-15x= 0 (The directions on my math packet say if the equation is not factor-able, use the quadratic formula to find the roots. What does this mean?)
Asked by Hayy-layyy :] - Mon Nov 3 00:40:17 2008 - - 6 Answers - 0 Comments
How do you solve these type quadratic Equations?
Q. Can someone please help me solve: e^2x + 3e^x - 10 The e is the constant e which is equivalent to 2.718281828. I need to see the working out not just the answer!!! You need to use the quadratic equation somehow???
Asked by BT is a legend - Wed Nov 11 04:30:06 2009 - - 6 Answers - 0 Comments
A. If =0, Assume e^x=y then , y^2+3y-10=0 (y+5)(y-2)=0 y=-5,e^x=-5 has no solution y=2, then , x ln e=ln 2, x=ln 2 God bless you.
Answered by nozar nazari - Wed Nov 11 04:38:25 2009
Q. Can someone please help me solve: e^2x + 3e^x - 10 The e is the constant e which is equivalent to 2.718281828. I need to see the working out not just the answer!!! You need to use the quadratic equation somehow???
Asked by BT is a legend - Wed Nov 11 04:30:06 2009 - - 6 Answers - 0 Comments
A. If =0, Assume e^x=y then , y^2+3y-10=0 (y+5)(y-2)=0 y=-5,e^x=-5 has no solution y=2, then , x ln e=ln 2, x=ln 2 God bless you.
Answered by nozar nazari - Wed Nov 11 04:38:25 2009
From Yahoo Answer Search: 'quadratic equations'
Tue Nov 17 19:12:41 2009 [ refresh local cache ]
[Hide]▼
news.google.com
Chronicle of Higher Education (subscription) (subscription)
Pearson promises future modules on quadratic equations , rationals, and several other topics. Students can even use the iPhone to buy paper textbooks online. ...
and more »
mathally.geekpedia.com
67px x 89px | 2.00kB
[source page]
We can write the final result
mathhelpforum.com
jzellt
ue, 10 Nov 2009 07:24:29 GM
Expand it out and you'll get a . quadratic equation. in x. Treat the a's as constants. Thread Tools. Posting Rules. You may not post new threads. You may not post replies. You may not post attachments. You may not edit your posts ...
[Hide]▲
