How to write a quadratic equation with integral coefficients such that one of its roots is 4-5i?
Q. How to write a quadratic equation with integral coefficients such that one of its roots is 4-5i? Help me! Oh My God!
Asked by Oops! - Wed Nov 5 20:45:09 2008 - - 2 Answers - 0 Comments

A. if 4-5i is one of the roots, you know its conjugate, 4+5i, also has to be a root. so you know (x-(4-5i))(x-(4+5i))=0, because that's how you would get roots from factoring, right? so anyway, if you multiply that mess out, you get: x^2 - 8x + 41 = 0
Answered by chickenwangs blemblemblem - Wed Nov 5 20:53:05 2008

When might it be better to solve a quadratic equation by factoring the polynomial without first?
Q. When might it be better to solve a quadratic equation by factoring the polynomial without first putting it in standard form? Not too sure. Perfect square form? or is that even a form?
Asked by JohnD - Fri Dec 14 22:04:38 2007 - - 1 Answers - 0 Comments

A. I would say that you would generally want to put in in standard form: ax2+bx+ c, first no matter what the situation. It would make factoring easier and defines the constants a,b,&c for the quadratic equation.
Answered by kennyk - Fri Dec 14 22:19:25 2007

What is the code for the quadratic equation on the ti-84 calculator and how do i store formulas?
Q. Im looking for the following codes, ive been searching everywhere on the internet 1) Quadratic Equation Code that also solves for nonreal numbers 2) is there any way to delete programs with out reformatting the whole calculator 3)If i wanted to say store all of the formulas for like triangles, trapezoids etc. how would i do that
Asked by sidd0123 - Tue May 19 19:17:50 2009 - - 1 Answers - 0 Comments

A. 1) A quadratic formula program for nonreal numbers is pretty easy; here's one that I made myself (and it only take two lines!): PrgmQUADFORM :Prompt A,B,C :Pause (-B+{1,-1} (B +4ACi ))/(2A Frac A few notes about this one: it displays decimals in fraction form, i.e. 1/3 instead of 0.333... Also, if the solution is longer than what can fit in the screen, you can scroll left and right to view the full length. 2) Of course! You didn't think Texas Instruments would leave out something so convenient, would they? Go to [2nd][+](Mem) > 2:Mem Mgmt/Del... > 7:Prgm... and find the ones you want to delete. When you've found it, hit the [Del] key. Select 'Yes' if it asks for confirmation. 3) You could always get an app like NoteFolio* to write notes… [cont.]
Answered by unknown - Tue May 19 21:14:49 2009

How to find the percentage using a quadratic equation?
Q. If you invested $12,000, part of it at 7% and the rest at 9% and made $970 in interest. How much was invested at 7%? I do you put that in a quadratic equation and solve it out?
Asked by Pookey - Tue Apr 28 10:33:14 2009 - - 1 Answers - 0 Comments

A. I assume the investment is for one year, and that the interest is paid yearly. Let x denote the amount (in dollars) invested at 7% Then (12000 - x) is the amount invested at 9% The investment at 7% pays 0.07x dollars interest The investment at 9% pays 0.09(12000 - x) dollars interest These two add to 970: 0.07x + 0.09(12000 - x) = 970 0.07x + 1080 - 0.09x = 970 -0.02x = -110 x = -110/-0.02 = 5500 Therefore, $5500 was invested at 7% and $6500 was invested at 9%. BTW, it's linear, not quadratic.
Answered by Ron W - Tue Apr 28 12:41:24 2009

How do I create a quadratic equation for the Sydney Harbor bridge? Please HELP!! ?
Q. In my Alg.2 class we are given the assignment to use quadratic equations in real life, our group chose the Sydney Harbor bridge. We found a few measurements, but how do we create the original equation? It is just to be given to us? Thanks!
Asked by Bailey - Fri Nov 28 09:41:38 2008 - - 1 Answers - 0 Comments

A. I don't know where this particular bridge is or what it looks like. BUT if it is in the shape of a parabola, it's arc will be a quadratic equation. All you need to do is have the measurement of the vertex. Imagine that the beginning of the bridge is on the origin (0,0) of a graph. Can you find the measure of the greatest height of the bridge? That would be the y-coordinate of the vertex. Then find the distance from one end to the vertex. That would be the x-coordinate of the vertex. Use the form: y - k = a(x-h) , putting the coordinates of the vertex in for (h,k). You'll need another point (length, height) to determine "a". Enjoy! ;)
Answered by Marley K - Fri Nov 28 10:08:55 2008

How do I use a quadratic equation to find the dimensions of a rectangle?
Q. The perimeter of the rectangle is 36 cm. Its area is 72 cm squared. How do I use a quadratic equation to find the dimensions of this rectangle?
Asked by jramirez23 - Sat Aug 9 20:12:06 2008 - - 4 Answers - 0 Comments

A. First you write a system. If x and y are the rectangle sides, you have: 2x+2y = 36 and xy = 72 x+y = 18 xy= 72 Find y in first equation and put it in the second: y= 18-x and x(18-x) = 72 and now you have the quadratic equation x^2 -18x +72 = 0 The roots are: 12 and 6 Then you have the answer: x=12 and y =6 ... it is the same rectangle if x=12 and y = 6
Answered by vahucel - Sat Aug 9 20:25:56 2008

How would I use the Discriminant to Solve this Quadratic Equation?
Q. For this Question, use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions. A designer attempts to arrange the characters of his artwork in the form of a square grid with equal numbers of rows and columns, but finds that 24 characters are left out. When he tries to add one more row and column, he finds that he has 25 too few characters. Find the number of characters used by the designer.
Asked by Cupake - Tue Nov 4 22:26:09 2008 - - 1 Answers - 0 Comments

A. the discriminant is the square root part in the quadratic equation solve that and if the answer is over zero its two solutions. equal to zero is one solution less than is no solution
Answered by Alexa A - Tue Nov 4 22:31:15 2008

Is it possible for a quadratic equation to have one real root and one imaginary root?
Q. Is it possible for a quadratic equation to have one real root and one imaginary root? I have to explain that question using examples. I know that its impossible but Im not sure what examples to use. (I'm on the math unit called 'quadratic functions..where we complete the square, factor, rationalization, discriminant...etc) What examples can I use? I have to use the discriminant formula i think.
Asked by unknown - Fri Apr 24 03:18:15 2009 - - 5 Answers - 0 Comments

A. THis is possible in case the coefficients are not real. for example (x+1)(x-i) = 0 or x^2 + (1-i)x - i = 0 but in case coefficients are real then it is not possible. In that case both shall be real or both complex comjugates .
Answered by mein Hoon na - Fri Apr 24 03:25:25 2009

How do I solve this quadratic equation by completing the square and applying the square root property?
Q. Solve the quadratic equation by completing the square and applying the square root property: 2a^2 + 8a 5 = 0 How do I do this? thanks so much for your help!!!
Asked by DoWHATiDO - Mon Nov 19 21:39:59 2007 - - 11 Answers - 0 Comments

A. [11] 2a^2+8a-5=0 2a^2+8a=5 a^2+4a=5/2[dividing both sides by 2] a^2+2*a*2+(2)^2=13/2 [adding (2)^2 or 4 to both sides] (a+2)^2=13/2 a+2=+-sqrt(13/2) [square-rooting both sides] a=-2+-sqrt(13/2)
Answered by alpha - Mon Nov 19 21:48:32 2007

How can You determine Roots from a quadratic Equation?
Q. Can any demonstrate an example of how you can determine the roots from a quadratic equation.
Asked by Charles C - Tue May 12 01:43:41 2009 - - 4 Answers - 0 Comments

How do I use the discriminant to determine the number of solutions of a quadratic equation?
Q. I am to use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex and it is not necessary to find the roots; just determine the number and types of solutions. This is the formula b^2 - 4ac, I'm supposed to use, but the 4 over 3x^2 - 2x + 3 over 4 again is completely confusing me and I have no idea how to start this. 4/3x^2 - 2x + 3/4 = 0
Asked by keltic1one - Sat Feb 14 20:23:45 2009 - - 4 Answers - 0 Comments

A. As for using the discriminant to determine the roots of a quadratic equation the rule is: b^2 - 4ac > 0 ... two real roots b^2 - 4ac = 0 ... one real root b^2 - 4ac < 0 ... two complex roots (which are conjugates of each other) In your example the discriminant is 2^2 - 4(4/3)(3/4) = 0 so one real root.
Answered by agent177 - Sat Feb 14 20:32:18 2009

How do I turn this into a quadratic equation?
Q. I have this question in my finance and I have been assured that this can be turned into a quadratic equation. Can someone please show me how to turn this into a quadratic equation. 0= -100 + 230/(1+x) - 132/(1+x^2) how can I make this into this format ax^2+bx+c = 0 Thanks everyone
Asked by J - Wed Feb 18 17:09:06 2009 - - 1 Answers - 0 Comments

A. You can turn it into a cubic equation if you multiply both sides by (1+x)(1+x^2), but you get a genuine cubic, not a quadratic: after doing the algebra you get 0 = -100x^3 + 130x^2 - 232x - 2 Which could be solved in terms of yucky radicals or something, but does _not_ factor nicely into a linear term and a quadratic term. If it were (1+x)^2 instead of (1+x^2) in the third term there [note the difference about where the square is], you could turn it into a quadratic by multiplying both sides by (1+x)^2. Then you'd get, after a little labor, 0 = -100x^2 + 30x - 2 which is indeed a genuine quadratic equation. But as written, no, it's not happening, despite whatever assurances you were given. :) Hope this helped
Answered by mcbengt - Thu Feb 19 23:31:49 2009

What's the best way to solve a quadratic equation?
Q. We're having a test on the quadratic equation, and I need to know what you think is the best and easiest way to solve one of these annoying problems. Also, are there any other methods? Oh yeah, and there's a bonus question on making your own quadratic equation using completing the square. Anyone know how to do this?
Asked by Kern - Wed Jun 6 01:10:05 2007 - - 21 Answers - 0 Comments

A. i think the best way to solve a quadratic equation is by factorising it (if possible). eg1. x^2 + 6x + 5 = 0 (x + 1)(x + 5) = 0 so x = -1 or x= -5 Or if you are allowed a calculator then you could just use the quadratic formula. ax^2 + bx + c = 0 x= (-b plus or minus the square root of ( (-b)^2 - 4(a)(c) ) ) all divided by 2a completing the square is also easy if the coefficient of x squared is 1. If not, you should probably use the formula..
Answered by Sally - Sun Jun 10 13:14:10 2007

when would factoring not work to solve a quadratic equation?
Q. I an find the zeros, roots, or solutions to a quadratic by taking the square root of both sides of the equation (when there is not a bx term), factoring, and completing the square.
Asked by frannieogreen - Wed Nov 19 21:41:55 2008 - - 3 Answers - 0 Comments

A. When the roots are irrational. This happens whenever the sqrt of the discriminant is irrational. In other words whenever b^2 - 4ac is not the square of a rational number.
Answered by xtempore - Wed Nov 19 21:45:27 2008

Would you ever use the quadratic equation when composing?
Q. Because in my honors algebra 2 class, we're going to have to do a report on a job where you use the quadratic equation. I've heard something about composing a piece of music that translates into a graph, or something like that, so I was wondering if this would work. I don't even know the quadratic equation yet, so sorry if this sounds really stupid.
Asked by Clare - Mon Feb 2 23:09:32 2009 - - 7 Answers - 0 Comments

A. There is a relationship between mathematics and music. After all, Pythagoras himself basically invented the study of acoustics when he realized that strings of different lengths would vibrate at different pitches when plucked. But you're going to see very little if not zero algebra in music. Music itself is very mathematical, but more in a geometric sense than an algebraic sense. The quadratic formula is x = negative b plus or minus the square root of b squared minus 4 a c, all divided by 2 a. Simply put, it just can't really apply to music. Music, particularly 20th century classical music has often been graphical in nature (composers would often create their own pictures or symbols in score in order to achieve a desired sound),… [cont.]
Answered by Ryan K - Mon Feb 2 23:27:03 2009

2. What three techniques can be used to solve a quadratic equation?
Q. What three techniques can be used to solve a quadratic equation? Demonstrate these techniques on the equation "x2 - 10x - 39 = 0".
Asked by ty b - Sun Mar 15 10:15:37 2009 - - 3 Answers - 0 Comments

A. You can factorise, complete the square or use the quadratic formula. (1) Factorising:- You need to get x -10x - 39 = 0 in the form (x + a)(x + b) = 0 So to do this quickly, you need two numbers which add to make -10 and multiply to make -39. You can tell the bigger of the two will be negative, and it works out to be:- (x - 13)(x + 3) = 0 And the roots of the equation, i.e. the values of x that make the equation true, will be when either (x - 13) = 0 or (x + 3) = 0. So the roots in this case are x = 13 or x = -3. (2) Completing the square:- Use the standard form: (x + a) = x + 2ax + a = 0 For your equation, x -10x - 39 = 0, we need 2a to equal -10, so this means a = -5. Substitute a = -5 into the above standard form - this… [cont.]
Answered by Oms at the Proms [Deletion] - Sun Mar 15 10:43:05 2009

Can someone give me an example of a word problem for a quadratic equation?
Q. I need to write about how to recognize a quadratic equation in the form of a word problem. I know I've worked with them before but I can't think of any for my project. Thanks for any help.
Asked by Dex - Sat May 10 19:08:38 2008 - - 2 Answers - 0 Comments

A. well, is this what you mean (and im just making this up off the top of my head, im not sure if it works, but ill try)? the first of three consecutive integers multiplied by the largest is 1 more than 4 times the middle. find the integers. the equation would look like: let x= first number x + 2=second number x + 4= third x(x+4)=4(x+2)+1 x +4x=4x+8+1 4x cancels x =9 x = 9 x=3 so the numbers would be 3, 5 and 7 hey, what do you know? it works. when you recognize a quadratic equation, it will always have something times itself. in this case, x *( x+4) is that. sometimes, you might have something like: the area of a square is 64. find the side length youd know that the area would be x =64 and that x would equal 8, and thats another example.… [cont.]
Answered by :) - Sat May 10 19:15:04 2008

10 POINTS - Write and solve a quadratic equation to determine when the cannon ball will reach sea level?
Q. A cannon ball is shot from a cannon on top of a hill that is 73.5 meters above sea level. The cannon ball leaves the cannon with a velocity of 9.8 m/s. Write and solve a quadratic equation to determine when the cannon ball will reach sea level.
Asked by Elliot J - Tue Aug 11 13:09:56 2009 - - 5 Answers - 0 Comments

A. -4.9t^2+73.5=0 -4.9 t^2 =-73.5 t^2=73.5/4.9 = 15 t=sqrt(15)=3.87 seconds
Answered by cidyah - Tue Aug 11 13:17:00 2009

Is it possible to solve a quadratic equation as a percentage of x?
Q. Example: 5x^2-7x-3=0 Is it possible to solve this so that x equals a number in %. I'm terrible at math and when I solve this equation I get a number for x1 and a number for x2. Is it possible to get just a number for x? Background behind the question: If the initial equation is: -1000/1+r + 600/(1+r)2=0, the answer for r=3.5 (350%). I hope thats right. But when the equation is -1000/1+r + 600/(1+r)2 + 1000/(1+r)3 = 0, then you get r1=.344 and r2=-1.744. This is for a homework assignment and the initial question asks for the rate of return for 2 years (the 350%), but the next part of the question adds another year and thats where I get confused. The "hint" from the teacher says "solve a quadratic equation in r when the equation =0. I' [cont.]
Asked by dizzamn - Sat Dec 2 23:40:25 2006 - - 3 Answers - 0 Comments

A. i'm confused too. when you solve 5x^2-7x-3=0, then you have two possible answers for x, and both are valid, that's how quadratic equations work. which one is right? you need more info. if you're doing compounded interest, and the initial amount is $100, rate of return is say 5%, then after 7 years you end up with 100*1.05^7=100*1.4071=$14 0.71 dollars after 7 years. now, is it possible to have a negative rate of return on a savings account? probably not! so you disregard the r2=-1.744 answer.
Answered by Nick C - Sat Dec 2 23:51:43 2006

Finding a quadratic equation that passes through certain points?
Q. I'm having a little trouble with this question: "Write an equation in the form y=ax^2+bx+c for the quadratic function whose graph passes through (7,0),(0,7) and (-1,0)". Looking at other questions I can kind of see how I'd do it but I have no idea if it would work out like that with these numbers. If anyone could take me through this, that would help a lot.
Asked by Gregory F - Sat Oct 25 16:19:53 2008 - - 3 Answers - 0 Comments

A. Well, we immediately know the value for 'c'. It must be 7, as demonstrated when we plug in the second point (0,7). y=ax^2+bx+7 (the value of '0' negates the 'ax^2' and 'bx' terms, so c=7) We can now set up two equations, using the other points: 0=a(-1)^2 +b(-1) +7 0=a(7)^2 +b(7) + 7 Simply solve this set of linear equations for 'a' and 'b'. a-b=-7 49a+7b=-7 56a=-56 a=-1, b=6, c=7
Answered by BoogyMan Messiah - Sat Oct 25 16:32:50 2008