Friday

April 18, 2014

April 18, 2014

Number of results: 214,232

**Math**

By the way, if you get one complex or imaginary root, you get two. They come in pairs, like male and female twins. One has +imaginary number, the other has -the same imaginary number. They are called complex conjugates. a + b i and a - b i and a can be zero.
*Wednesday, February 20, 2008 at 11:55am by Damon*

**math**

imaginary as in mecosonic, just changing the order of the letters but more than likely it's not a word you'll see in the dictionary, that's what real or imaginary refers to.
*Friday, July 16, 2010 at 3:20pm by unknown*

**11th grade math**

Just add the real and imaginary terms separately, and keep them separate. The imaginary terms are the ones with "i" -2 + 3i -1 - 21 -------- -3 + i
*Monday, January 4, 2010 at 12:01am by drwls*

**Algebra**

State the possible number of imaginary zeros of g(x)= x^4+3x^3+7x^2-6x-13 this is how far Ive gotten:there are 4 zeros,3 positive and 1 negative If you've found four real zeroes, then there are no imaginary zeroes. You can also see that there are no imaginary zeroes directly: ...
*Friday, August 10, 2007 at 5:13pm by Marissa*

**Math**

Which describes the number and type of roots of the equation x^2-625=0? A)1 real root, 1 imaginary root B)2 real roots, 2 imaginary roots C)2 real roots D)4 real roots I went with A sqrt x^2= x (which would be the imaginary) and sqrt 625= 25 (the real)
*Wednesday, February 20, 2008 at 11:55am by Jon*

**Precalculus**

"Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the two real roots and the four imaginary roots of this equation." I used synthetic division to get the real roots 2 and -1, but I can't figure out how to get the imaginary roots. I checked HotMath but all it showed...
*Sunday, September 20, 2009 at 10:51pm by Emily*

**algebra ll-imaginary numbers**

where can i find hepl online about imaginary numbers? i do not understand them.
*Wednesday, March 18, 2009 at 7:50am by coral*

**Precalculus**

how about this let y = x^3 then your equation becomes y^2 - 7y - 8 = 0 (y-8)(y+1) = 0 so y = 8 or y = -1 then x^3 = 8 or x^3 = -1 x^3 - 8 = 0 (x-2)(x^2 + 2x + 4) = 0 this will give you the real root of x=2 and 2 imaginary roots from the quadratic or x^3 + 1 = 0 (x+1)(x^2 - x...
*Sunday, September 20, 2009 at 10:51pm by Reiny*

**math**

i is an imaginary number. YOu can graph it only on the complex number plane. Consider a rectangular graph, the x axis is the real numbers, and the y axis is imaginary numbers. THen you plot the point (-3,4). Find the distance from the origin, it is 5.
*Tuesday, February 2, 2010 at 10:23am by bobpursley*

**Algebra II**

Also, I think I confused you above by saying "there are no real solutions, but there are 2 imaginary solution". I should have just said 2 imaginary solutions, since that was what you were asking. sorry to confuse you.
*Saturday, January 8, 2011 at 5:33pm by helper*

**Math**

-sqrt47 is not imaginary. It is real, and irrational. sqrt(-47) is imaginary. (2ef-3*g2/3e-2 * f4 * g)-2 first do the inner.. (2/3 * e1+2f-3+4g2+1)-2
*Wednesday, November 5, 2008 at 5:06pm by bobpursley*

**Algebra**

The function 5x^2+2x-1's domain is all real numbers, according to the answer key. However, I don't know how to factor this to prove it. Could you show me? The function the sq. root of y-10 has a domain of all numbers so that y is greater than or equal to zero. But I don't see ...
*Sunday, November 1, 2009 at 12:19pm by Anneliese*

**Algebra**

The function f(x)=x^4−10x^3+40x^2−80x+64 has four complex roots, one of which is 2−2i. What is the sum of all real and imaginary coefficients of these roots? i here is imaginary unit i.e. i^2 = -1
*Monday, March 18, 2013 at 3:22am by Stranger*

**alg. 2**

Because -81^3 is a negative number. If you take a 1/4 root of any negative number (which is the same thing as taking a square root twice), you get imaginary or complex (real + imaginary) numbers.
*Friday, March 7, 2008 at 5:48pm by drwls*

**maths2**

Formula for discriminant: d = b^2 - 4ac note that if d = 0 : real, equal/double root d < 1 : imaginary roots d > 1 : real, unequal roots 7x2 + 3x + 1 = 0 d = 3^2 - 4(7)(1) d = 9 - 28 d = -19 Thus, it's D. imaginary roots. Hope this helps~ :3
*Monday, October 7, 2013 at 1:32pm by Jai*

**Imaginary land Homework poster**

create a 2-D or 3 -D imaginary land.Land must include at least 5 of the following landforms:volcano,mountain range,plateau,valley,island,lake,!@#$%^&ula,plain,and river.to create a SNAZZY name
*Thursday, November 10, 2011 at 2:29pm by gracie*

**Math**

Hi, I have to write a research paper and I decided to write about the history/importance of imaginary numbers. I don't know how to get started though, can you please help ! By the way I know how to write a research paper so don't give me the links about how to write 1. Instead...
*Friday, April 1, 2011 at 11:07pm by Jon*

**Maths**

The function f(x)=x^4−15(x^3)+81(x^2)−201x+182 has four complex roots, one of which is 3−2i. What is the sum of all real and imaginary coefficients of these roots? Details and assumptions i is the imaginary unit, where i2=−1.
*Tuesday, April 16, 2013 at 3:33am by OIan*

**Maths**

Suppose z=a+bi, where a and b are integers and i is the imaginary unit. We are given that |1+iz|=|1−iz| and |z−(13+15i)|<17. Find the largest possible value of a+b. i is the imaginary unit, where i^2=−1.
*Friday, April 5, 2013 at 8:43pm by Oian*

**maths**

Suppose z=a+bi, where a and b are integers and i is the imaginary unit. We are given that |1+iz|=|1−iz| and |z−(13+15i)|<17. Find the largest possible value of a+b. Details and assumptions i is the imaginary unit, where i^2=−1.
*Wednesday, April 3, 2013 at 4:37am by rohit*

**Physics**

The magnitude or "mod" is the square root of the sum of the squares of real and imaginary terms. The phase angle in the arctangent of the ratio (imaginary part/real part). Most of this was explained in the link that I gave you earlier. BobPursley is more experienced with AC ...
*Saturday, March 29, 2008 at 6:37am by drwls*

**algebra**

Any "real" number multiplied by itself is positive. Therefore there is no "real" square root of a negative nunber. That does not mean that you cannot define imaginary numbers that have the property that their squares are negative. Imaginary numbers are an important part of ...
*Monday, June 30, 2008 at 10:51pm by drwls*

**algebra2-imaginary numbers**

You can add or subtract complex numbers by treating the i as a variable and combining like terms. I am having a lot of trouble figuring out these equations with imaginary numbers. 1. (3+2i)+(7-i)= 2.(1-6i)+(2-i)= 3. (2+i)-(3+i)= 4.(4+i)-(2-i)= Can someone please walk me ...
*Tuesday, January 16, 2007 at 3:52pm by chrissy*

**Imaginary land Homework poster**

I'd use homemade clay to create a 3-D land. Sketch your imaginary land out on paper first. Decide which landforms you're going to show and where you'll put them. Then, using a piece of plywood or other sturdy backing, sculpt your imaginary land. As you're working, I'm sure you...
*Thursday, November 10, 2011 at 2:29pm by Ms. Sue*

**algebra**

Hi. I answered you previous post and it was almost similar, except that here, the roots must be complex/imaginary so D must be less than zero. Anyway, Recall the formula for discriminant. For a quadratic equation in the general form, ax^2 + bx + c = 0, D = b^2 - 4ac if D = 0...
*Tuesday, October 8, 2013 at 12:52pm by Jai*

**Math**

I suggest you start by doing lots of research and reading. Take good notes. http://www.google.com/#hl=en&sugexp=llsfp&pq=ip%20address%20lookup&xhr=t&q=imaginary+numbers&cp=17&pf=p&sclient=psy&aq=0&aqi=&aql=&oq=imaginary+numbers&pbx=1&bav=on.2,or.r_gc.r_pw.&fp=71a4fa0fe339f87d
*Friday, April 1, 2011 at 11:07pm by Ms. Sue*

**Algebra II**

Substitute x = i t where t is assumed to be real: t^4 - 3it^3 - 7t^2 -6 i t - 13 = 0 Take the imaginary part of the equation: t^3 + 2 t = 0 ---> t(t^2 + 2) = 0 Now, we can see from the original equation for t that t = 0 is not a solution (this is actually implied by the ...
*Sunday, February 17, 2008 at 11:45am by Count Iblis*

**math**

my method works for any odd-numbered square draw the square, add an imaginary row to the top and an imaginary column at the right start with 1 in the middle of the top row and and place consecutive numbers following a righ diagonal pattern. If you reach the imaginary top row, ...
*Wednesday, March 13, 2013 at 8:52pm by Reiny*

**math**

Imaginary numbers.
*Tuesday, January 5, 2010 at 8:12pm by Mysterio *

**math**

isn't it an imaginary number
*Tuesday, April 13, 2010 at 4:56pm by pem*

**Math**

Do you get 1 b/c of the x in front of the x itself? ( imaginary 1)
*Wednesday, February 22, 2012 at 6:26pm by Cassie*

**math**

by the way you cant factor x^2+ 1 with imaginary i's i believe
*Friday, February 1, 2013 at 9:25pm by vii*

**pre-calc**

It multiplies both the real and imaginary parts by the same real factor. This increases the length r in the polar vector (r, theta) representation, but not its angle, theta. Any complex number can be written x + iy or r e^(i*theta) where r cos theta is the real part, x , and r...
*Thursday, February 21, 2008 at 8:02am by drwls*

**Math**

im not sure, but i think that this is imaginary.. no solution...
*Sunday, January 13, 2008 at 9:24pm by EAW*

**Math**

(5-4i)/(8+i) ~ simplify the expression (i=imaginary #) help plz! thanks!
*Saturday, December 27, 2008 at 2:01pm by Naoko*

**Math**

find all real and imaginary solutions x^(2/3)-2=x^(1/3)
*Thursday, October 21, 2010 at 10:52pm by Samantha*

**algebra**

question The square root of a negative number is imaginary. That is you cannot find, for example, sqrt(-49) . Why? my answer Yes you can, however this is aginst the normal operations non teh less we are dealing with iminiganary numbers so, as such the square root of a ...
*Monday, June 30, 2008 at 10:51pm by tony*

**Math - Imaginary Numbers**

Remember that i represents the square root of (-1). So, i^2 = -1 i^5 = (i^2)(i^2)(i) =(-1)(-1)(i) =?
*Thursday, September 11, 2008 at 6:41pm by Quidditch*

**Math - Imaginary Numbers**

Remember that i^2 = -1 3(-2i)^2 =3(-2)^2(i^2) =? -i(isqrt(5))^2 =-i(i^2)(sqrt(5))^2 =?
*Thursday, September 11, 2008 at 6:41pm by Quidditch*

**math**

How do I write 5e^(-2j) in real and imaginary forms?
*Tuesday, October 14, 2008 at 3:36pm by steve*

**Math**

sqrt49 is Rational. sqrt(-81) = 9i, Imaginary.
*Monday, October 31, 2011 at 4:35pm by Henry*

**math**

x^3 - 16x = 0 x(x^2 - 16) = 0 x(x+4)(x-4) = 0 x = 0 , 4 , -4 there are 3 real roots, none imaginary.
*Monday, January 16, 2012 at 9:31am by Reiny*

**Algebra**

3 i or 3 times the square root of -1 If you have not had imaginary numbers in your course, the answer is no real number solution no, square root of 2 or [ sqrt 2 } Yes there is an answer to the square root of a negative number but not in real numbers. If you have not covered ...
*Saturday, September 8, 2012 at 8:38pm by Damon*

**Trig (first one)**

If -2 cosx -i sin y = 2 sinx +i both the real and imaginary terms on opposite sides of the equation must be equal. -2 cos x = 2 sin x Divide both sides by 2 cos x to solve for x -1 = tan x x = 3 pi/4 and 7 pi/4 From the imaginary part of the equation sin y = -1 y = (3/2) pi
*Friday, May 30, 2008 at 5:54pm by drwls*

**Math(I see that now)**

I made it harder than it was thanks. But I just thought that x was imaginary
*Wednesday, February 20, 2008 at 11:55am by Jon*

**Math - Imaginary Numbers**

I just need help with these questions, thanks. i^5 3(-2i)^2 -i(isqrt5)^2
*Thursday, September 11, 2008 at 6:41pm by Lucy*

**math**

solve the equation for all real and imaginary solutions (x-3/2)^2-2(x-3/2)+26=0
*Thursday, October 21, 2010 at 11:32pm by Allie*

**Algebra II-Reiny-Please check**

Above I stated I had worked this problem out and there were two IMAGINARY solutions but I'm still confused about the question which asks how many solutions arethere to the problem-should I put "0" or do the two imaginary solutions count and I should put "2". That's what I'm ...
*Saturday, January 8, 2011 at 5:33pm by Matt C*

**Math**

Find all real or imaginary solutins to each equation. 3y(2)+4v-1=0
*Thursday, July 31, 2008 at 12:09am by Marilyn*

**math**

I am not familiar with the term to "work" an equation. Are you trying to solve for x, or for i? Is i an imaginary number?
*Sunday, August 8, 2010 at 1:39pm by Reiny*

**math**

Find all real and Imaginary solutions 7x^(3)+70x^(2)-x-10=0
*Tuesday, October 26, 2010 at 11:00am by Samantha*

**math**

find the imaginary solutions of the equation by factoring. x to the third power - 16x = 0
*Monday, January 16, 2012 at 9:31am by Ryan*

**Maths**

We want to find two real numbers p and q such that (p+qi)(p+qi)=(5+12i). Collect the real and the imaginary terms together into two separate equations: Real terms: p²-q²=5 Imaginary terms: 2pq=12, i.e. pq=6 You could solve the above as a pair of simultaneous equations, but ...
*Thursday, September 18, 2008 at 5:41pm by David Q*

**Algebra**

x^2 = -10 x = ±√-10 = ±i√10 two imaginary or complex roots. (in this case it was easier to find the actual roots when determining the "nature of the roots". the value of b^2 - 4ac would have been 0 - 4(1)(10) or -40 , so you would have two imaginary roots)
*Sunday, September 4, 2011 at 8:37pm by Reiny*

**Algebra II**

The answer would be 2. If the question had been worded how many 'real solutions' (which it isn't) your answer would be 0, since the answer is an 'imaginary number' not 'real number'. Do you understand the difference between real number (solutions) and imaginary number (...
*Saturday, January 8, 2011 at 5:33pm by helper*

**math**

how many different 9-letter words (real or imaginary) can be formed from the letters in the word ECONOMICS?
*Friday, July 16, 2010 at 3:20pm by sha*

**Algebra**

No. You just inspect the function (in this case (-3x+1) and see if there are any values of x that would violate the fundamental laws of math. For instance, if you had (-3x+1)/(x+3) then the domain would be all values of x EXCEPT x=-3, as division by zero is NOT allowed. In ...
*Friday, May 7, 2010 at 11:37am by bobpursley*

**Math**

imaginary numbers Find the reciprocal of 2-7i and express it in terms of a+bi I need help please and thank you
*Monday, March 12, 2012 at 6:31pm by Anonymous*

**Math**

Find all real or imaginary solutions to each equation. Use the method of your choice. 3y² + 4v – 1 = 0
*Wednesday, July 30, 2008 at 10:38pm by Marilyn*

**math (trig)**

Im having problems solving this equation: 2sin(x) = cos(x) + 2 I keep getting an imaginary number..:/
*Tuesday, December 18, 2007 at 8:02pm by jess*

**math**

Sorry, I did not notice that is was an equation. I was just rewriting the right hand side. x would be the imaginary number + or - i sqrt15
*Wednesday, March 3, 2010 at 9:28pm by drwls*

**Math ( Pre Calc)**

Find all real and imaginary roots of the polynomial equation 3x^4-x^3+4x^2-2x-4=0
*Tuesday, October 2, 2012 at 1:06pm by Ains*

**Algebra II**

Which describes the number and type of roots of the equation x^2 -625=0? A. 1 real root, 1 imaginary root B. 2 real roots, 2 imaginary roots C. 2 real roots D. 4 real roots. I have x^2 = 625 x = 25 answer: 2 real roots (25 or -25) Is this correct? Thanks
*Sunday, March 16, 2008 at 5:24pm by Lucy*

**Math**

The general equation for the function is: f( x ) = a ( x - x1 ) ( x - x2 )( x - x3 ) where x1, x2 and x3 are the roots and a is the coefficient of x ^ 3 There is a theorem in algebra called the Imaginary Zeros Theorem. It says that for a polynomial that is restricted to real ...
*Friday, May 24, 2013 at 5:49am by Bosnian*

**math**

how many diferrent 9-letter words (real or imaginary) can be formed from the letters in the word ECONOMICS? the ans. must be 90,720
*Sunday, July 18, 2010 at 6:12pm by sha*

**10th Grade Math**

Why is is important to understand addition? The same reasoning applies, exponents (rational, real, irrational, or imaginary) are basic operations in mathematics. If you want to be somebody, using math logic, you need to be expert in them.
*Sunday, September 21, 2008 at 3:33pm by bobpursley*

**Math**

your calculator will surely revolt at your attempt try: 4 x √ -1 = mine says "Error 2" Calculators usually are not programmed to do imaginary numbers
*Friday, January 18, 2013 at 11:42am by Reiny*

**algebra**

What are the intercepting points of y^2=-4x and x^2=y I set both equations equal to 0, but only get x=0 and there is another intercepting point. please help! Graphically, y=x^2 is positive for all values of x, whether positive or negative. The second equation is y positive for...
*Thursday, December 7, 2006 at 4:48pm by mary*

**MATH**

How can you factor x^4 +1? If you go to the imaginary plane, it factors to (x^2 + i)(x^2-i) I need to factor by difference of squares. How can I do this?
*Sunday, March 25, 2007 at 4:01pm by Mike*

**math**

The imaginary number i is defined such that i2 = –1. What does i + i2 + i3 + ... + i23 equal?
*Friday, September 24, 2010 at 5:50pm by Cennie*

**math**

b^2 - 4ac = 441 - 4(49)(9/2) = -441 which is less than zero , so there are two imaginary roots
*Saturday, March 23, 2013 at 11:58am by Reiny*

**math**

. Use the discriminant to determine whether the following equations have solutions that are: two different rational solutions; two different irrational solutions; exactly one rational solution; or two different imaginary solutions. 8x2 + 2x + 4 = 0 Two different irrational ...
*Monday, May 2, 2011 at 11:26am by carol*

**Math**

Solve: (don't forget about imaginary numbers) 5x^2-7x+12=0 how do I solve this? please help and thank you
*Monday, March 12, 2012 at 5:27pm by Anonymous*

**math**

There is a solution. The solution is an imaginary number.
*Wednesday, November 11, 2009 at 9:23pm by Nick *

**math**

A^2 = A squared A^2 + 7^2 = 6^2 A^2 + 49 = 36 A^2 = -13 Getting the square root of both sides would give you an imaginary number. Do you have a typo in this problem?
*Wednesday, December 2, 2009 at 7:15pm by PsyDAG*

**math**

Hello, I'd like to know how in developing, factoring or simplifying, we can move from: goo.gl/zZdzSs ACDE are strictly positive real constants. I is an imaginary number.
*Saturday, November 2, 2013 at 6:03pm by Karl*

**algebra,math,**

is this factorable its a trinomial: x^2+2x+3 No, not factorable in the real number system. b^2 -4ac is negative, so the roots are imaginary.
*Wednesday, April 18, 2007 at 9:31pm by jacky20*

**math**

solve and simplify the answer completely by using pure imaginary numbers/complex numbers. 2x^2+3x+2=0
*Thursday, December 4, 2008 at 9:20pm by anonymous*

**math**

find whether the line 2x-y=0 is tangent, real chord on imaginary chord to the parabola y^2-2y+4y=0.
*Monday, August 8, 2011 at 10:08am by sonu*

**Math**

you dont need to add dots at the end of the zero on the 1000000000000000000. There is always an imaginary zero beside every number.
*Monday, March 29, 2010 at 8:07pm by anonymous*

**math**

Hello, i would like to know what is the complex number of : (a*c*d)×(-e*d+i)/(1-e*c*d^2) ACDE are strictly positive real constants. I is an imaginary number. Thanks you a lot for you help.
*Saturday, November 2, 2013 at 9:09pm by Karl*

**Math 111**

Solve the equation by the square root property. If possible, simplify radicals or rationalize denominators. Express imaginary solutions in the form a+bi. [X+ 2/3]^2=10/9
*Friday, April 2, 2010 at 6:46pm by Anonymous*

**Calculus III**

Express sin(3x) in terms of sin(x). exp(ix) = cos(x) + i sin(x) (1) Take the third power of both sides: exp(3ix) = [cos(x) + i sin(x)]^3 From (1) it follows that the imaginary part of the left hand side is sin(3x) Expand the right hand side and take the imaginary part. You ...
*Thursday, March 13, 2008 at 2:43pm by Count Iblis*

**Math**

Hi, I have to write a research paper and I decided to write about the history/importance of imaginary numbers. I don't know how to get started though, can you please help ! :) thanks
*Friday, April 1, 2011 at 5:32pm by Jon*

**MATH**

USING AM IMAGINARY FIGURE OF $19,250.00 AS AN ANNUAL INCOME FIGURE, HOW MUCH MONEY COULD YOU SAVE A YEAR IF YOU BUDGETED HALF OF 2.4%?
*Wednesday, September 7, 2011 at 5:50pm by J*

**math**

Solve: (don't forget about imaginary numbers) 5x^2-7x+12=0 how do I solve this? do I need to use the quadratic formula? please help and thank you
*Monday, March 12, 2012 at 7:18pm by Anonymous*

**Math**

Hello, I'd like to know how in developing, factoring or simplifying, we can move from: A + ( (1 / ( (1 / ICD) + IED) ) to: ACD *( ( - ED + I ) / (1 - ECD^2 ) ) ACDE are strictly positive real constants. I is an imaginary number.
*Saturday, November 2, 2013 at 4:07pm by Karl*

**Physics PLEASE**

imaginary number =_+
*Tuesday, June 23, 2009 at 1:54pm by HELP ME*

**Algebra B**

imaginary number
*Tuesday, February 1, 2011 at 8:19pm by helper*

**math**

You have a difference of squares (x^2-3)(x^2+3) One of these factors to two real roots, the other to two imaginary roots. You have a difference of squares (x^2-3)(x^2+3) One of these factors to two real roots, the other to two imaginary roots. Let g(x)= x^4 - 9 wat are the ...
*Wednesday, May 30, 2007 at 9:22pm by bobpursley*

**math**

18x^3 + 15x^2 + 18x = 0 3x(6x^2 + 5x + 6) = 0 so x = 0 or x = ...... (use the quadratic formula, I see imaginary roots coming up here )
*Monday, November 15, 2010 at 1:58pm by Reiny*

**math**

Hello, I'd like to know how in developing, factoring or simplifying, we can move from: A + ( (1 / ( (1 / ICD) + IED) ) to: ACD *( ( - ED + I ) / (1 - ECD^2 ) ) ACDE are strictly positive real constants. I is an imaginary number. Thanks a lot for your help.
*Saturday, November 2, 2013 at 10:14am by Karl*

**math**

nine letters, two sets of two. 9!/2!2! check my thinking. I still am in awe at what an imaginary word is....must be something in new age poetry.
*Friday, July 16, 2010 at 3:20pm by bobpursley*

**math**

(fog)(x) = f(g(x)) = [g(x)]^2 (fog)(x) = [sqrt(x-3)]^2 note that the sqrt and ^2 will just cancel each other out: (fog)(x) = x - 3 to find its domain,, recall that domain is the set of all possible values of x,, since we both considered f(x) and g(x) here, we find the ...
*Thursday, March 31, 2011 at 6:49am by Jai*

**HCA 220**

Yes it has to be an imaginary patient
*Wednesday, January 20, 2010 at 5:22pm by Isabel*

**Trigonometry**

They must be imaginary, I don't see them
*Sunday, December 8, 2013 at 10:57pm by Reiny*

**Trigonometry**

They must be imaginary, I don't see them
*Sunday, December 8, 2013 at 10:57pm by Reiny*

**math**

Why conjugage? The conjugage of a given expression is obtained by changing the sign of the imaginary part. For example, the conjugate of(5x+4i) is (5x-4i), or the conjugate of (2-3i) is (2+3i). By multiplying the numerator and denominator by the complex conjugate, the ...
*Wednesday, June 10, 2009 at 3:12pm by MathMate*

**Maths**

I assume those 1/2 's are exponents so your question is x = √(2+√5) + √(2-√5) y = √(2+√5) - √(2-√5) Now clearly √(2-√5) is not a real number , so this must deal with imaginary numbers. for ease of typing let a = &#...
*Sunday, September 16, 2012 at 11:49am by Reiny*

**Algebra**

Yes, the answer is D There is also an imaginary solution x = 0, y = +/- i
*Thursday, February 21, 2008 at 1:54pm by drwls*

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