quadratic equation it has 2 answers . can someone give me step by step solutions?
929526
4x^2 - 16 = 0\n4(x^2 - 4) = 0\nx^2 - 4 = 0\n(x - 2)(x + 2) = 0\n\nANS : x = 2 or -2
534396
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.
215050
2x²-2x-3=0\n2(x²-x-3/2)=0\nx²-x-3/2=0\n\napply the quadratic formula\n=>a=1 b=-1 c=-3/2\n\n=>[-(-1) + or - √(-1)²-4(1)(-3/2) ]over 2(1)\n\n=>[1 + or - √1+6] over 2\n=>[1 + or - √7] over 2 [answer]
1172065
if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
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(
371573
If i is a root, you may express it into x = i, a root of x^3 + i = 0.\nsince x = i, x - i = 0\nTherefore, x^3 + i has a root x - i. to get the other 2 factors, divide.\n(x^3 + i)/(x - i) = 0\nDivide it using long division, and you'll get\nx^2 + ix - 1 = 0\nActually, \nx^3 + i factors as (x - i)(x^2 + ix - 1) \ntherefore, \nx^2 + ix - 1 = 0\nuse the quadratic formula:\na = 1\nb = i\nc = -1\nx = [-b +or- sqrt(b^2 - 4ac)]/2a\nx = [-i +or- sqrt(i^2 - 4(1)(-1))]/2(1)\nx = [-i +or- sqrt(-1 + 4)/2\nx = (-i +or- sqrt -3)/2\nx = (-i +or- i sqrt3)/2\nTherefore, the other roots are\nx = (-i + i sqrt3)/2\nx = (-i - i sqrt3)/2\n\n^_^
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
961776
That is not the quadratic formula it is -b plus or minus, b squared - 4(ac) all over 2a, you would best to just look up the quadratic formula to understand it, you can also use factorization, or completeing the square, but must have a coefficient of one.  For larger pol nomials there are other ways, but just best to google it or yahoo it considering the site i am on.
1168578
a) If x(x+2) = 0, then either term x or (x+2) must equal zero.  If either term is zero, then multiplying it times the other term, or anything else, would yield a product of zero as well.\n\nSo, if x = 0, then x(x+2) = 0 (0 + 2) = 0 * 2 = 0\nor if x = -2, then x(x+2) = -2 (-2 + 2) = -2 * 0 = 0\n\nTherefore, x = 0, -2\n\n\nb) more work to be done here.  Multiply the terms and get zero on one side of the equation.\n\ny(y+2) = 3\ny^2 + 2y - 3 = 3 - 3\ny^2 + 2y - 3 = 0\n\nNow, factor the quadratic equation.\n\n(y + 3)(y - 1) = 0  (because if you multiply the two terms on the left side, you get y^2 + 2y - 3).\n\ny values of -3 and 1 will make the equation = 0\n\n(-3 + 3)(-3 - 1) = 0 * -4 = 0\n(1 + 3)(1 - 1) = 4 * 0 = 0\n\nSo, y = -3, 1
791337
1) (1/2 i+2bj) - (bi-b^2 j) + ((7/2 +b)i-bj) = 4i+20j\n\nNow separate the equation into two equations.\n\n1/2 i - bi + (7/2 + b)i = 4 i\n2bj + b^2 j - bj = 20j\n\ni and j are no longer needed so get rid of them too\n\n1/2 - b + 7/2 + b = 4 => 0 = 0 (Not a big help)\n2b + b^2 - b = 20 \n\nb^2 + b - 20 = 0\n\nTime for the quadratic formula.\n\nb = (-1 +- sqrt(1 + 80)) / 2 = -1/2 +- 9/2 = -5, 4\n\n2) x = 1/2 i+2bj\n\nBasically you need to find where b makes the length of the vector equal to one. The Pythagorean Theorem helps here.\n\n(1/2)^2 + (2b)^2 = 1^2\nb = sqrt(3/4) / 2
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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.
316686
the process u follow is correct, but actually it is wrong. fundamentally wat u did is ok, but technically it is wrong. because this becomes a quadratic equation which has more than 1 root. so this equation has more than 1 answers for which it is true.
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
870960
Do you mean that the roots of x^2+pX+r are α  and β and you need to find quadratics in terms of p and q?\nIf so you can use the following:\nIn any quadratic ax^2 + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a.\nUsing this,  -(α + β) = p, and αβ= r.\n\nAnswers are for terms b and c in the quadratic x^2 + bx + c\ni)\nb = -(1/α + 1/β) = -(α + β)/αβ = p/r\nc = 1/αβ = 1/r\n\nii) \nb = -(1/α^2 + 1/β^2) =  -(α^2 + β^2)/(αβ)^2 = -[(α + β)^2 - 2αβ])/(αβ)^2 = -(p^2 - 2r)/r^2\nc = 1/(αβ)^2 = 1/r^2\n\niii) \nb = (α^2β + αβ^2) = αβ(α + β) = -pr\nc = -α^2β * -αβ^2 = (αβ)^3 = r^3
265248
the general form of a quadratic equation ax^2 + bx + c =0\n\ntherefore there is two solution\nx=[-b ± sqrt (b^2-4ac)]/(2a)\n\nx^2+2x+2=0\n\na=1 b=2 c=2\n\nx1=[-2+sqrt(2^2-4*1*2)]/2 =(-2+2i)/2 = -1+i\nx2=[-2-sqrt(2^2-4*1*2)]/2 =(-2-2i)/2 = -1-i\n\nwhere i is the square root of -1 and it is imaginary\n\nthe solution is X={-1+i , -1-i}\n\nThose solution means that this equation in the from y=x^2+2x+2 has no intersection with y=0 since no real solution for x^2+2x+2=0