How am I supposed to solve for this algebraic equation?
827429
letus take all terms on LHS to RHS SO WE WILL GET\n23x^2-16x+2x+3-18=0\n 23x^2-14x-15=0\n \n\nhere a=23 b=(-14)  c=(-15)\n find D= b^2-4ac\n\nthen x={ (-b)+(square root of D)}/2a\n\nor x={ (-b)-(square root of D)}/2a
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
979866
If I am reading this as you have submitted, then, the solution of x=2secx is transcendental, and needs to be solved graphically giving a set of values that satisfy it; you plug that set of values for x that satisfy it into y=3+2tanx, and that gives the solution set for y.
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)
928663
Are you asking for help or an answer?\n\nTake any 2 equations and multiply all of the terms or one or both by numbers that will make one of the terms in each equation have the same coefficient. Now subtract one from the other and the term will disappear. Now take the resulting equation and do the same with it against the third. Keep doing this until you are down to one term and solve for the variable. Now substitute the variable into the equation and solve for the other variables.
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
681679
The solution involves complex numbers ('i' notation)\n\nThe factorized equation is:\n\n(b + 3 - i)(b + 3 + i)\n\nwhere 'i' is the imaginary complex number = square root of -1\n\n(TOOK ME WHOLE OF 15 MINS!!! MY MATH IS SLIPPING!!!)\n\neinsteinbrainiac - you came close to the solution
770289
It's the difference between two perfect squares.  Take the square root of 16 and x squared.  Then a minus sign and the square root of 25.  Put parentheses around all of that. Write it again, and change the minus sign to a plus sign.  You will set each group in parentheses equal to zero.  There will be 2 equations.  Solve for x.  X will have 2 solutions.
832845
you need to add +1\nso that the equation can be represented as a perfect square\nx^2 - 2x + 1 = (x - 1) (x - 1) = (x - 1)^2\n\nthis is called solving by perfecting the square\nto do this, make sure that the coefficient of the first term is 1.\nThen, divide the coefficient of the middle term by 2,\nthen square the whole thing.\n\nIn this case, the coefficient of the middle term is -2\n-2 / 2 = -1\n(-1)^2 = +1
151739
algebra is the solution of equations like the above.  Let the "what" be expressed as w.  Then the question becomes\n\n  w * 3 = 4/3\n\nNow how do we solve this for w.  The trick is to operate on both sides of the equation the same function.  The caveat is that the function applied must have an inverse.  In this case multiply both sides by (1/3).\n\n(1/3)*3*w = (1/3)*(4/3)\n\nNow since 1/3 is multiplicative inverse of 3, the result is 1, and since 1 is the multiplicative identity, the result of the left hand side is w.  For the right hand side just do the multiplication\n\nw = 4/9\n\nThus 4/9 multiplied by 3 equals 4/3.
134065
Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
197473
This is not an equation unless it is equal to something.  It is an expression.\n\nFirst, you can simplify the middle: a-5a = -4a\n\nThen, multiply the first 2 together: 3a(-4a)=-12a^2\n\nNow multiply this by the last part: -12a^2(a-35)= -12a^3 + 420a^2.\n\nIf you were supposed to set this equal to 0 and solve for a, you would not need to multiply it out.  You would just set each part = 0 and solve\n\n3a = 0\n-4a = 0\na-35 = 0
583062
i'm not sure if i know what you're asking for, but to solve a quadratic equation you should just be able to use your solve( command and enter in the expression.  be sure to put a comma and the variable name at the end of it, for example \n\nsolve(x^2 + 2x +3=0, x)\n\np.s. i just typed that in randomly, it may have imaginary roots, which your calculator can handle but you'll have to use csolve(
865391
I think there is a sign error: the cubic term is supposed to be -2x^3, not 2x^3. Then the other roots are 3 and -1 and the factors are\n\n(x-8i)(x+8i)(x-3)(x+1)
982041
Well dear\n\n-If the first root is " α " and the second roots is " α + 3"  so have  \n▪x ² - (α + β) + α β = 0 \n▪( x - α )(x - β ) = 0 \n-b/a = α+ β  \n&\nc/a = α β\nso we have ; \nc/a = m/4  \n-b/a = - ( -16) / 4 = 4 \nα + α + 3 = 4 , 2α + 3 = 4 \nα = 1/2  ,  β = 3 1/2 \n2 /a = α β  =>  m/4 =  1/2 * 7/2  , m = 7\n\nGood Luck dear