What are the steps for solving quadratic equations/inequalities?
Q. The quadratic equations I am given to solve are: -8x^2+48x-40 = 0 and -8x^2+48x-40 > 0. I just need a reminder on how to solve them. Thanks. :) Need to solve these: -8x^2+48x-40 = 0, -8x^2+48x-40 > 0, -8x^2+48x-40 = 15.
Asked by Hayy-layyy :] - Sun Nov 9 23:27:38 2008 - - 4 Answers - 0 Comments
A. 1) -8x^2 + 48x - 40 = 0 (-8x^2 + 48x - 40)/-8 = 0/-8 x^2 - 6x + 5 = 0 x^2 - x - 5x + 5 = 0 (x^2 - x) - (5x - 5) = 0 x(x - 1) - 5(x - 1) = 0 (x - 1)(x - 5) = 0 x - 1 = 0 x = 1 x - 5 = 0 x = 5 x = 1 , 5 = = = = = = = = 2) -8x^2 + 48x - 40 > 0 -8(x^2 - 6x + 5) > 0 x^2 - 6x + 5 < 0/-8 x^2 - 6x + 5 < 0 (x - 1)(x - 5) < 0 x - 1 > 0 and x - 5 < 0 x - 1 > 0 x > 1 x - 5 < 0 x < 5 x > 1 and x < 5
Answered by intellectual - Mon Nov 10 00:16:36 2008
Q. The quadratic equations I am given to solve are: -8x^2+48x-40 = 0 and -8x^2+48x-40 > 0. I just need a reminder on how to solve them. Thanks. :) Need to solve these: -8x^2+48x-40 = 0, -8x^2+48x-40 > 0, -8x^2+48x-40 = 15.
Asked by Hayy-layyy :] - Sun Nov 9 23:27:38 2008 - - 4 Answers - 0 Comments
A. 1) -8x^2 + 48x - 40 = 0 (-8x^2 + 48x - 40)/-8 = 0/-8 x^2 - 6x + 5 = 0 x^2 - x - 5x + 5 = 0 (x^2 - x) - (5x - 5) = 0 x(x - 1) - 5(x - 1) = 0 (x - 1)(x - 5) = 0 x - 1 = 0 x = 1 x - 5 = 0 x = 5 x = 1 , 5 = = = = = = = = 2) -8x^2 + 48x - 40 > 0 -8(x^2 - 6x + 5) > 0 x^2 - 6x + 5 < 0/-8 x^2 - 6x + 5 < 0 (x - 1)(x - 5) < 0 x - 1 > 0 and x - 5 < 0 x - 1 > 0 x > 1 x - 5 < 0 x < 5 x > 1 and x < 5
Answered by intellectual - Mon Nov 10 00:16:36 2008
How do you use a graphing calculator in solving quadratic equations?
Q. For example, 27x^4 + 43.2x^3 / x^2 - 3.7x + 3.22 = 1.8 x 10^(-7) I need the answer specifically using the calculator. It's for equilibrium in chemistry. I've think I know how to solve for the quadratic equation when it equals zero, but how do I do it when it doesn't equal zero (like in my example). I'm using a TI-84
Asked by Anne - Tue Mar 18 23:47:29 2008 - - 3 Answers - 0 Comments
A. You can solve any equation by graphing the LHS (left hand side) and RHS (you guess!) separately (as Y1 and Y2 on many calculators), and then finding the intersections... The x-value(s) of the intersection(s) will be the solutions to the equation. For your example, you could graph (27x^4 + 43.2x^3)/(x^2 - 3.7x + 3.22) as Y1, and 1.8 x 10^(-7) as Y2, and find the intersecion(s). By the way... 1.8x 10^(-7) is zero, practically speaking!
Answered by notthejake - Tue Mar 18 23:53:34 2008
Q. For example, 27x^4 + 43.2x^3 / x^2 - 3.7x + 3.22 = 1.8 x 10^(-7) I need the answer specifically using the calculator. It's for equilibrium in chemistry. I've think I know how to solve for the quadratic equation when it equals zero, but how do I do it when it doesn't equal zero (like in my example). I'm using a TI-84
Asked by Anne - Tue Mar 18 23:47:29 2008 - - 3 Answers - 0 Comments
A. You can solve any equation by graphing the LHS (left hand side) and RHS (you guess!) separately (as Y1 and Y2 on many calculators), and then finding the intersections... The x-value(s) of the intersection(s) will be the solutions to the equation. For your example, you could graph (27x^4 + 43.2x^3)/(x^2 - 3.7x + 3.22) as Y1, and 1.8 x 10^(-7) as Y2, and find the intersecion(s). By the way... 1.8x 10^(-7) is zero, practically speaking!
Answered by notthejake - Tue Mar 18 23:53:34 2008
algebra 1 help on solving quadratic equations ten points for best answer?
Q. Using square roots to solve quadratic equations Algebra 1? solve the equation. Round the solutions to the nearest hundredth. 1. 12-xsquared = 17 2. 9- x squared = 9 3. 8xsquared = 50 4. ( x- 3) squared = 5 5. ( x+2)squared = 10 6.3 ( x- 4)squared = 18 please give me an explanation on how to solve these. Thankyou and God bless.
Asked by just chill chill just chill - Mon May 5 18:19:06 2008 - - 2 Answers - 0 Comments
A. 1) 12 from both sides , and multiply by (-1) this gives x2=-5 then if you take the square root of both sides you get x = sqroot of -5 2) you get zero , (9-9 =0 ) 3)so : if you divide both sides by 8 , you get x2 = 50/8 then you just take the square root of that . x = sqroot of 50/8 or 2.5 4) foiling this out , you get x2 -6x+ 9 = 5 , then subtract 9 to get x2-6x=-4 factor out an x , you get x(x-6)=-4 set each part equal to -4 , you get x=-4 , x-6=-4 which then gives x=2 so your answers are x=-4 , x=2 5) same process as #4 you get x=6 x=2 6) same process as 4&5 you get : x2-8x=2 x=2 x=10 you're very welcome and God bless you too
Answered by rarabear21 - Mon May 5 19:37:36 2008
Q. Using square roots to solve quadratic equations Algebra 1? solve the equation. Round the solutions to the nearest hundredth. 1. 12-xsquared = 17 2. 9- x squared = 9 3. 8xsquared = 50 4. ( x- 3) squared = 5 5. ( x+2)squared = 10 6.3 ( x- 4)squared = 18 please give me an explanation on how to solve these. Thankyou and God bless.
Asked by just chill chill just chill - Mon May 5 18:19:06 2008 - - 2 Answers - 0 Comments
A. 1) 12 from both sides , and multiply by (-1) this gives x2=-5 then if you take the square root of both sides you get x = sqroot of -5 2) you get zero , (9-9 =0 ) 3)so : if you divide both sides by 8 , you get x2 = 50/8 then you just take the square root of that . x = sqroot of 50/8 or 2.5 4) foiling this out , you get x2 -6x+ 9 = 5 , then subtract 9 to get x2-6x=-4 factor out an x , you get x(x-6)=-4 set each part equal to -4 , you get x=-4 , x-6=-4 which then gives x=2 so your answers are x=-4 , x=2 5) same process as #4 you get x=6 x=2 6) same process as 4&5 you get : x2-8x=2 x=2 x=10 you're very welcome and God bless you too
Answered by rarabear21 - Mon May 5 19:37:36 2008
What is the formula in solving quadratic equations by completing the square?
Q. I really need to have this formula mastered because I'm in for MTAP, and in case you don''t know that, it's a Mathematics Qualifying Examination. So, I sorta need this today-or I'm out...! HEEELLP!!!
Asked by jc_delrosario143 - Thu Jan 14 04:46:32 2010 - - 7 Answers - 0 Comments
A. This is an example of Completing the Square Question Number 1 : For this equation x^2 - 13*x + 36 = 0 , answer the following questions : A. Use completing the square to find the root of the equation ! Answer Number 1 : The equation x^2 - 13*x + 36 = 0 is already in a*x^2+b*x+c=0 form. As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = -13, c = 36. 1A. Use completing the square to find the root of the equation ! x^2 - 13*x + 36 = 0 ,divide both side with 1 Which result in x^2 - 13*x + 36 = 0 , Which means that the coefficient of x is -13 We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -13/2 = -6.5 Which means we can turn the equation into x^2 - 13*x + 42.2 [cont.]
Answered by seed of eternity - Thu Jan 14 08:52:00 2010
Q. I really need to have this formula mastered because I'm in for MTAP, and in case you don''t know that, it's a Mathematics Qualifying Examination. So, I sorta need this today-or I'm out...! HEEELLP!!!
Asked by jc_delrosario143 - Thu Jan 14 04:46:32 2010 - - 7 Answers - 0 Comments
A. This is an example of Completing the Square Question Number 1 : For this equation x^2 - 13*x + 36 = 0 , answer the following questions : A. Use completing the square to find the root of the equation ! Answer Number 1 : The equation x^2 - 13*x + 36 = 0 is already in a*x^2+b*x+c=0 form. As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = -13, c = 36. 1A. Use completing the square to find the root of the equation ! x^2 - 13*x + 36 = 0 ,divide both side with 1 Which result in x^2 - 13*x + 36 = 0 , Which means that the coefficient of x is -13 We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -13/2 = -6.5 Which means we can turn the equation into x^2 - 13*x + 42.2 [cont.]
Answered by seed of eternity - Thu Jan 14 08:52:00 2010
Can anyone please explain the proper method on "Solving Quadratic Equations by Extraction of Roots"?
Q. I'm a sophomore in a Philippine High School, our Math lesson right now is all about Quadratic Equations. Our teacher gave us a group project, in where we have to choose a particular method on solving a Quadratic Equation and discuss and present it in front of the class, but me and my groupmates hardly know how to solve using the topic that we got: "Solving Quadratics by Extraction of Roots". That is why I'm just wondering what the easy, sure and proper way on doing this method...
Asked by Jhong - Mon Sep 24 07:14:21 2007 - - 3 Answers - 1 Comments
A. A quadratic equation is of the form ax^2 + bx + c = 0, where a, b, and c are real constants and a is not zero. If a = 0, then it is no longer a quadratic equation. It, then, becomes a linear equation as the term ax^2 will then be zero. The equation can be written as a [ x^2 + (b/a) x + (c/a) ] =0 Now, using first two terms, make a perfect square. To do that we have to add and subtract (b/2a)^2. Thus, equation is a [ x^2 + (b/a) x + (b/2a)^2 - (b/2a)^2 + (c/a) ] = 0 => a [ { x + (b/2a) }^2 - (b^2 - 4ac) / 4a^2 ] = 0 => [ x + (b/2a) ]^2 = (b^2 - 4ac) / 4a^2 [ because a is not zero ] => x + (b/2a) = +/- sqrt (b^2 - 4ac) / 2a => x = - (b/2a) +/- sqrt (b^2 - 4ac) / 2a These are the roots of the quadratic… [cont.]
Answered by Madhukar Daftary - Mon Sep 24 07:49:13 2007
Q. I'm a sophomore in a Philippine High School, our Math lesson right now is all about Quadratic Equations. Our teacher gave us a group project, in where we have to choose a particular method on solving a Quadratic Equation and discuss and present it in front of the class, but me and my groupmates hardly know how to solve using the topic that we got: "Solving Quadratics by Extraction of Roots". That is why I'm just wondering what the easy, sure and proper way on doing this method...
Asked by Jhong - Mon Sep 24 07:14:21 2007 - - 3 Answers - 1 Comments
A. A quadratic equation is of the form ax^2 + bx + c = 0, where a, b, and c are real constants and a is not zero. If a = 0, then it is no longer a quadratic equation. It, then, becomes a linear equation as the term ax^2 will then be zero. The equation can be written as a [ x^2 + (b/a) x + (c/a) ] =0 Now, using first two terms, make a perfect square. To do that we have to add and subtract (b/2a)^2. Thus, equation is a [ x^2 + (b/a) x + (b/2a)^2 - (b/2a)^2 + (c/a) ] = 0 => a [ { x + (b/2a) }^2 - (b^2 - 4ac) / 4a^2 ] = 0 => [ x + (b/2a) ]^2 = (b^2 - 4ac) / 4a^2 [ because a is not zero ] => x + (b/2a) = +/- sqrt (b^2 - 4ac) / 2a => x = - (b/2a) +/- sqrt (b^2 - 4ac) / 2a These are the roots of the quadratic… [cont.]
Answered by Madhukar Daftary - Mon Sep 24 07:49:13 2007
What are the disadvantages in using completing the square for solving quadratic equations?
Q. What are the disadvantages in using completing the square for solving quadratic equations?
Asked by katia.antonova - Sun Jan 11 23:54:07 2009 - - 3 Answers - 0 Comments
A. It's generally more time consuming than factoring (if applicable) and using the quadratic formula. Here's an example x^2 + 6x + 8 = 0 Solving by completing the square: Add 9 to both sides. x^2 + 6x + 9 + 8 = 9 Factor the now-square trinomial. (x + 3)^2 + 8 = 9 (x + 3)^2 = 1 Square root both sides, x + 3 = +/- 1 x = -3 +/- 1 x = { -3 - 1, -3 + 1 } x = { -4, -2 } As opposed to factoring: x^2 + 6x + 8 = 0 (x + 4)(x + 2) = 0 x = { -4, -2 } Or the quadratic formula: x^2 + 6x + 8 = 0 x = [-6 +/- sqrt(6^2 - 4(1)(8)) ] / [ 2 ] x = [-6 +/- sqrt(36 - 32)]/2 x = [-6 +/- sqrt(4)]/2 x = [-6 +/- 2]/2 x = {(-6 + 2)/2 , (-6 - 2)/2} x = {-4/2, -8/2} x = {-2, -4} (In this case it looks to be more time consuming to do the quadratic formula) But… [cont.]
Answered by Puggy - Mon Jan 12 00:02:23 2009
Q. What are the disadvantages in using completing the square for solving quadratic equations?
Asked by katia.antonova - Sun Jan 11 23:54:07 2009 - - 3 Answers - 0 Comments
A. It's generally more time consuming than factoring (if applicable) and using the quadratic formula. Here's an example x^2 + 6x + 8 = 0 Solving by completing the square: Add 9 to both sides. x^2 + 6x + 9 + 8 = 9 Factor the now-square trinomial. (x + 3)^2 + 8 = 9 (x + 3)^2 = 1 Square root both sides, x + 3 = +/- 1 x = -3 +/- 1 x = { -3 - 1, -3 + 1 } x = { -4, -2 } As opposed to factoring: x^2 + 6x + 8 = 0 (x + 4)(x + 2) = 0 x = { -4, -2 } Or the quadratic formula: x^2 + 6x + 8 = 0 x = [-6 +/- sqrt(6^2 - 4(1)(8)) ] / [ 2 ] x = [-6 +/- sqrt(36 - 32)]/2 x = [-6 +/- sqrt(4)]/2 x = [-6 +/- 2]/2 x = {(-6 + 2)/2 , (-6 - 2)/2} x = {-4/2, -8/2} x = {-2, -4} (In this case it looks to be more time consuming to do the quadratic formula) But… [cont.]
Answered by Puggy - Mon Jan 12 00:02:23 2009
Solving quadratic equations with negative discriminant?
Q. I'm not lazy, I promise! We never learned this in class, and it's on my math exam. Please help? I'll be forever indebted to you =]. Solve for x using quadratic formula: 1. 5x^2-2x+2=0 2. 2x^2+8x+9=0 The answers given are: 1. (1 3i)/5 2. (4 i 2)/2 Thank you kindly xoxoxo.
Asked by Tomorrow-xo - Sat Jun 13 19:10:46 2009 - - 4 Answers - 0 Comments
A. The general form of a quadratic equation is :- ax^2 + bx + c = 0 the quadratic formula to obtain the roots of the equation is:- x = ( -b ( b^2 - 4ac ) ) / 2a you can plug in the values from the above equations into this formula. 1. a = 5, b = -2, c = 2 when put in the formula u get:- (2 (-36) )/ 10 here (-36) = 6i Thus solution is (1 3i)/5 2. a = 2, b = 8, c = 9 Similar process for this as well. remember a negative value in underoot gives a complex solution. ( -1) = i which is called IOTA.
Answered by kid rock - Sat Jun 13 19:23:45 2009
Q. I'm not lazy, I promise! We never learned this in class, and it's on my math exam. Please help? I'll be forever indebted to you =]. Solve for x using quadratic formula: 1. 5x^2-2x+2=0 2. 2x^2+8x+9=0 The answers given are: 1. (1 3i)/5 2. (4 i 2)/2 Thank you kindly xoxoxo.
Asked by Tomorrow-xo - Sat Jun 13 19:10:46 2009 - - 4 Answers - 0 Comments
A. The general form of a quadratic equation is :- ax^2 + bx + c = 0 the quadratic formula to obtain the roots of the equation is:- x = ( -b ( b^2 - 4ac ) ) / 2a you can plug in the values from the above equations into this formula. 1. a = 5, b = -2, c = 2 when put in the formula u get:- (2 (-36) )/ 10 here (-36) = 6i Thus solution is (1 3i)/5 2. a = 2, b = 8, c = 9 Similar process for this as well. remember a negative value in underoot gives a complex solution. ( -1) = i which is called IOTA.
Answered by kid rock - Sat Jun 13 19:23:45 2009
when solving quadratic equations, how do you know what method to use?
Q. for example when a question says prove that this equation equals zero would you use the quadratic equation or would you solve it by completing the square or by factorising? Anyone who can help me? I would be very grateful!!! Thanks x
Asked by Lisa A - Fri Jun 1 09:18:09 2007 - - 5 Answers - 0 Comments
A. I tend to use either eyeballing or the quadratic equation. I mean if you have something like x^2 - 2x - 24 you can look at it and see that this is (x - 6)(x + 4). If I can't just see the roots in about 10 seconds, I use the quadratic eqution. I don't like completing the square, because I think there's too much potential to be working with weird fractions. With the quadratic equation, you don't have to deal with that sort of stuff until the very end. But it's been about twenty-five years since I was taught all this stuff, and we were never taught to complete the square, so maybe I'm just unfamiliar with it.
Answered by TychaBrahe - Fri Jun 1 09:24:18 2007
Q. for example when a question says prove that this equation equals zero would you use the quadratic equation or would you solve it by completing the square or by factorising? Anyone who can help me? I would be very grateful!!! Thanks x
Asked by Lisa A - Fri Jun 1 09:18:09 2007 - - 5 Answers - 0 Comments
A. I tend to use either eyeballing or the quadratic equation. I mean if you have something like x^2 - 2x - 24 you can look at it and see that this is (x - 6)(x + 4). If I can't just see the roots in about 10 seconds, I use the quadratic eqution. I don't like completing the square, because I think there's too much potential to be working with weird fractions. With the quadratic equation, you don't have to deal with that sort of stuff until the very end. But it's been about twenty-five years since I was taught all this stuff, and we were never taught to complete the square, so maybe I'm just unfamiliar with it.
Answered by TychaBrahe - Fri Jun 1 09:24:18 2007
Help with Solving Quadratic Equations by extracting the square root?
Q. This is the problem: 2x = 8/(2x + 7) clarification: the numbers on the left side means 8 over 2x + 7. Can you show the solution step-by-step? Thanks.
Asked by Phantom^Phreak - Tue Oct 6 06:15:54 2009 - - 6 Answers - 0 Comments
A. 2x = 8/(2x+7) 2x(2x+7)=8 4x^2+14x-8=0 x = [-14+{14^2-4(4)(-8)}^(1/2 )]/2(4) x = [-14 + 18]/8 x = 1/2 x = [-14-{14^2-4(4)(-8)}^(1/2 )]/2(4) x = [-14 - 18]/8 x = -4 x = 1/2 or x = -4
Answered by Prabh Pal Singh - Tue Oct 6 06:31:04 2009
Q. This is the problem: 2x = 8/(2x + 7) clarification: the numbers on the left side means 8 over 2x + 7. Can you show the solution step-by-step? Thanks.
Asked by Phantom^Phreak - Tue Oct 6 06:15:54 2009 - - 6 Answers - 0 Comments
A. 2x = 8/(2x+7) 2x(2x+7)=8 4x^2+14x-8=0 x = [-14+{14^2-4(4)(-8)}^(1/2 )]/2(4) x = [-14 + 18]/8 x = 1/2 x = [-14-{14^2-4(4)(-8)}^(1/2 )]/2(4) x = [-14 - 18]/8 x = -4 x = 1/2 or x = -4
Answered by Prabh Pal Singh - Tue Oct 6 06:31:04 2009
Solving quadratic equations by completing the square?
Q. could somone show me how to solve this equations by completing the square? 5=X -2x I also need to know the roots
Asked by MrSquiggles - Sun Apr 19 15:56:04 2009 - - 2 Answers - 0 Comments
Q. could somone show me how to solve this equations by completing the square? 5=X -2x I also need to know the roots
Asked by MrSquiggles - Sun Apr 19 15:56:04 2009 - - 2 Answers - 0 Comments
Solving Quadratic equations by using the quadratic formula?
Q. can you show me how to do this, step by step; b^2+11b+30=0 using x = -b+or- the square root of b^2-4ac all over 2a thanks,, i was just making sure i got the right answer, because my friend got a completely different answer from me.
Asked by Makayla S - Wed May 20 23:52:46 2009 - - 2 Answers - 0 Comments
Q. can you show me how to do this, step by step; b^2+11b+30=0 using x = -b+or- the square root of b^2-4ac all over 2a thanks,, i was just making sure i got the right answer, because my friend got a completely different answer from me.
Asked by Makayla S - Wed May 20 23:52:46 2009 - - 2 Answers - 0 Comments
Who can help me with two simple math problems? Solving quadratic equations?
Q. Solve by the square root method. 4-3y^2=-20 Solve by factoring. 15x^2-120x=0 Please explain. I have a test tomorrow, lol. Thank you.
Asked by Sodapop :] - Mon Nov 16 21:54:11 2009 - - 2 Answers - 0 Comments
A. 4-3y^2=-20 -3y^2=-16 3y^2=16 (multiply both sides by -1) 3y^2= 16 3y=4 (square root of 16 is 4) 3y/3=4/3 y=4/3 And 15x^2-120x=0 x^2-8x=0 (I divided each side by 15) (x-8)(x+0)=0 therefore x=8 or x=0
Answered by chicken_fun7 - Mon Nov 16 21:59:57 2009
Q. Solve by the square root method. 4-3y^2=-20 Solve by factoring. 15x^2-120x=0 Please explain. I have a test tomorrow, lol. Thank you.
Asked by Sodapop :] - Mon Nov 16 21:54:11 2009 - - 2 Answers - 0 Comments
A. 4-3y^2=-20 -3y^2=-16 3y^2=16 (multiply both sides by -1) 3y^2= 16 3y=4 (square root of 16 is 4) 3y/3=4/3 y=4/3 And 15x^2-120x=0 x^2-8x=0 (I divided each side by 15) (x-8)(x+0)=0 therefore x=8 or x=0
Answered by chicken_fun7 - Mon Nov 16 21:59:57 2009
Solving quadratic equations with perfect squares?
Q. I need to know how to do this, i don't exactly know the steps of how to do this, so if you could help that would be great! 6x^2 - 18 = 0 Thanks!
Asked by [ily.] jonasbros.+michaeljackson - Tue Jan 20 17:36:52 2009 - - 2 Answers - 0 Comments
A. Add 18 to both sides 6x^2=18 Divide both sides by 6 x^2=18/6 Take the square root of both sides x=SRT3 x=-SRT3
Answered by Minty - Tue Jan 20 17:59:52 2009
Q. I need to know how to do this, i don't exactly know the steps of how to do this, so if you could help that would be great! 6x^2 - 18 = 0 Thanks!
Asked by [ily.] jonasbros.+michaeljackson - Tue Jan 20 17:36:52 2009 - - 2 Answers - 0 Comments
A. Add 18 to both sides 6x^2=18 Divide both sides by 6 x^2=18/6 Take the square root of both sides x=SRT3 x=-SRT3
Answered by Minty - Tue Jan 20 17:59:52 2009
how do i go about solving quadratic equations using alpha and beta formula?
Q. how do i go about solving quadratic equations using alpha and beta formula?
Asked by bintajay - Sun Aug 19 17:11:46 2007 - - 3 Answers - 0 Comments
A. No such formula.
Answered by ironduke8159 - Sun Aug 19 17:24:00 2007
Q. how do i go about solving quadratic equations using alpha and beta formula?
Asked by bintajay - Sun Aug 19 17:11:46 2007 - - 3 Answers - 0 Comments
A. No such formula.
Answered by ironduke8159 - Sun Aug 19 17:24:00 2007
When solving quadratic equations, what are the pros and cons of using factoring versus the quadratic formula?
Q. When solving quadratic equations, what are the pros and cons of using factoring versus the quadratic formula?
Asked by Jen R - Mon Apr 16 23:36:01 2007 - - 5 Answers - 0 Comments
A. Pros to factoring -sometimes it is very simple -sometimes it is very quick Cons to factoring -sometimes, it just wont work Pros to quadratic formula -always, always works -can be used in a calculator method to do it for you Cons to quadratic formula -takes a long time -you have to memorize stuff...
Answered by Grant A - Mon Apr 16 23:40:53 2007
Q. When solving quadratic equations, what are the pros and cons of using factoring versus the quadratic formula?
Asked by Jen R - Mon Apr 16 23:36:01 2007 - - 5 Answers - 0 Comments
A. Pros to factoring -sometimes it is very simple -sometimes it is very quick Cons to factoring -sometimes, it just wont work Pros to quadratic formula -always, always works -can be used in a calculator method to do it for you Cons to quadratic formula -takes a long time -you have to memorize stuff...
Answered by Grant A - Mon Apr 16 23:40:53 2007
Solving quadratic equations word problems?
Q. Sandra knits a sweater for her daughter; she has 6 rolls of yarn; she s completed 2 sides. How many rolls will she need to complete the sweater?
Asked by ladysoul2u2 - Mon Jun 15 21:47:43 2009 - - 1 Answers - 0 Comments
A. We need more information. Back from the days when I was knitting sweaters we only had two sides (actually a front and a back) and sleeves. From what you have given us we have no idea what portion of the sweater has been completed.
Answered by Julius N - Tue Jun 16 03:02:05 2009
Q. Sandra knits a sweater for her daughter; she has 6 rolls of yarn; she s completed 2 sides. How many rolls will she need to complete the sweater?
Asked by ladysoul2u2 - Mon Jun 15 21:47:43 2009 - - 1 Answers - 0 Comments
A. We need more information. Back from the days when I was knitting sweaters we only had two sides (actually a front and a back) and sleeves. From what you have given us we have no idea what portion of the sweater has been completed.
Answered by Julius N - Tue Jun 16 03:02:05 2009
solving Quadratic Equations by using the Quadratic Formula?
Q. 4m^2+3m=1
Asked by Kaila M - Tue Jul 22 09:41:55 2008 - - 2 Answers - 0 Comments
A. Quadratic Formula -b (b^2 - 4ac) ___ 2a Given: 4m^2 + 3m = 1 Remember: a = 4m^2, b = 3m, c = 1 1) 4m^2 + 3m -1 = 0 2) -3 [(3^2) - 4(4)(-1)] ___ 2(4) 3) -3 9 + 16 ___ 8 4) -3 25 ___ 8 5) -3 5 ___ 8 Possible answers: a) positive: 1/4 b) negative: -1
Answered by julimel.nathan - Tue Jul 22 10:07:32 2008
Q. 4m^2+3m=1
Asked by Kaila M - Tue Jul 22 09:41:55 2008 - - 2 Answers - 0 Comments
A. Quadratic Formula -b (b^2 - 4ac) ___ 2a Given: 4m^2 + 3m = 1 Remember: a = 4m^2, b = 3m, c = 1 1) 4m^2 + 3m -1 = 0 2) -3 [(3^2) - 4(4)(-1)] ___ 2(4) 3) -3 9 + 16 ___ 8 4) -3 25 ___ 8 5) -3 5 ___ 8 Possible answers: a) positive: 1/4 b) negative: -1
Answered by julimel.nathan - Tue Jul 22 10:07:32 2008
What is the British method of solving a quadratic equation?
Q. I skipped the school Algebra class(taught myself it over the summer so I could do 8th grade Geometry) and I heard them talking about the British method of solving quadratic equations or something like that. Could someone tell me what this is? Thank you!
Asked by Griffin Henderson - Tue Aug 26 15:35:42 2008 - - 2 Answers - 0 Comments
A. There is no difference. It is exactly the same.
Answered by cidyah - Tue Aug 26 17:10:44 2008
Q. I skipped the school Algebra class(taught myself it over the summer so I could do 8th grade Geometry) and I heard them talking about the British method of solving quadratic equations or something like that. Could someone tell me what this is? Thank you!
Asked by Griffin Henderson - Tue Aug 26 15:35:42 2008 - - 2 Answers - 0 Comments
A. There is no difference. It is exactly the same.
Answered by cidyah - Tue Aug 26 17:10:44 2008
Compare and Contrast four different methods of solving quadratic equation?
Q. Can you please help me compare and contrast the four different methods of solving quadratic equations? These different ways are graphing, quadratic formula, completing the square, and factoring.
Asked by sally - Sat Jan 10 23:01:06 2009 - - 1 Answers - 0 Comments
A. The quadratic formula is derived by completing the square of the general form of a quadratic equation, so these two methods are essentially the same method. (Any teacher who disagrees with this does not know what he/she is talking about.) A benefit to both is that they always provide an exact solution to any quadratic equation. A draw back to both is that they can be kind of cumbersome. Solving a quadratic by graphing it on a calculator (as opposed to sketching it by hand) is easy, but usually only provides the user with an approximation (i.e., you zoom in on the roots and find some decimal number that is close but not exact). Also, if the quadratic has imaginary roots this method will fail completely (unless you have a very expensive… [cont.]
Answered by Baker Street Irregular - Sun Jan 11 06:49:08 2009
Q. Can you please help me compare and contrast the four different methods of solving quadratic equations? These different ways are graphing, quadratic formula, completing the square, and factoring.
Asked by sally - Sat Jan 10 23:01:06 2009 - - 1 Answers - 0 Comments
A. The quadratic formula is derived by completing the square of the general form of a quadratic equation, so these two methods are essentially the same method. (Any teacher who disagrees with this does not know what he/she is talking about.) A benefit to both is that they always provide an exact solution to any quadratic equation. A draw back to both is that they can be kind of cumbersome. Solving a quadratic by graphing it on a calculator (as opposed to sketching it by hand) is easy, but usually only provides the user with an approximation (i.e., you zoom in on the roots and find some decimal number that is close but not exact). Also, if the quadratic has imaginary roots this method will fail completely (unless you have a very expensive… [cont.]
Answered by Baker Street Irregular - Sun Jan 11 06:49:08 2009
solving quadratic equations by factoring?
Q. I am trying to solve this following equation by factoring: 15 = 8x(squared) - 14x. i am not sure how to, please help!
Asked by Kimi - Wed Aug 5 18:37:35 2009 - - 1 Answers - 0 Comments
Q. I am trying to solve this following equation by factoring: 15 = 8x(squared) - 14x. i am not sure how to, please help!
Asked by Kimi - Wed Aug 5 18:37:35 2009 - - 1 Answers - 0 Comments
From Yahoo Answer Search: 'solving quadratic equations'
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gmating.com
Wed, 04 Mar 2009 19:40:14 GM
the roots of this . equation. is given by. -2--- x = - b- - . This formula can be used to . solve. all . quadratic equations. in the form ax2 + bx + c = 0 with a = 0 . . Solve. x2 - 2x + 2 = 0 . Solution. We put a = 1, b = 2 and c ...
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