If the value of b^2 - 4ac is negative, the equation ax^2 + bx + c = 0?
953700
The discriminant\n\n√b² - 4ac portion of the quadratic Formula\n\nx = - b ± √b² - 4ac/ 2a is negative , no real number solutions \n\n- - - - - -s-
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
340490
(ax^2 + bx + c)(dx^2 + ex + f)\n=ax^2(dx^2 + ex + f) + bx(dx^2 + ex + f) + c(dx^2 + ex + f)\n=adx^4 + aex^3 + afx^2 + bdx^3 + bex^2 + bfx + cdx^2 + cex + f\n=adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + f
496516
It depends on the signs of the terms (a,b in the example).\n\n(ax + b) ^ 2 = (ax)^2 + 2abx + b^2\n\nthe middle term is 2ab, so if a and b have different signs, it will be negative\n\nLike this:\n\nthe square of :\n(2x + 3) is ( 4x^2 + 12x + 9 )\n(2x - 3) is ( 4x^2 - 12x + 9 )\n(-2x + 3) is ( 4x^2 - 12x + 9 )\n(-2x - 3) is ( 4x^2 + 12x + 9 )
785246
x^2-16=0\n(x+4)(x-4)=0\nx+4=0\nx=-4\n\nx-4=0\nx=-4\n\nx^3+14x^2+40x=0\nx(x^2-14x+40)=0\nx(x-10)(x-4)=0\nx=0\n\nx-10=0\nx=10\n\nx-4=0\nx=4\n\nx^4-5x^2+4=0\n(x^2-4)(x^2-1)=0\n(x-2)(x+2)(x-1)(x+1)=0\nxi2=0\nx=2\nx+2=0\nx=-2\nx-1=0\nx=1\nx+1=0\nx=-1
911422
The discriminant is b^2 - 4ac. If it looks familiar, it's in the quadratic formula.\n\nThe quadratic is in the form ax^2 + bx + c. Plug in your values for a, b, and c.\n\nIf your result is:\n....positive, there are two real solutions\n...zero, there is one real solution (that is repeated)\n...negative, there are two imaginary solutions.\n\nHope that helped.
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
193529
if your equation is of the form\nax^2 +bx+c=0 then the general formula for the solution is\nthe two roots are:\n(-b+(b^2-4ac)^(1/2))/(2a)\n(-b-(b^2-4ac)^(1/2))/(2a)\nif b^2-4ac is less than zero then you will have imaginary roots, in such a case ur roots would be\n(-b+i(4ac-b^2)^(1/2))/(2a)\n(-b-i(4ac-b^2)^(1/2))/(2a)
1156185
You know the first term is x^2 so you can write:\n(x     )(x     )\nYou know that the second term is -4x:\nIf factors are of the form (x+A)(x+B) then  Ax+Bx = -4x\nYou also know that the last term is -5, so A*B=-5\n\nJust find the values of A and B that satisfy these equations:\nAx+Bx = -4x\nA*B = -5\n\nA = -5 and B = 1\n\nNOW that we have the factors\n(x-5)(x+1)=0\nTo get 0 either section of the left hand side can be 0, so solve both cases:\nCASE 1:\nx-5 =0\nx = 5\nCASE 2:\nx+1=0\nx=-1\n\nANSWER x=5 or x=-1
781653
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
258820
4x^2 -3x + 2 = 0 \nD = b^2 - 4ac = 9 - 4*2*4= 9- 32 = - 23<0\nThere are no real solutions, but there are 2 complex solutions.
749429
x = -7\n\nMultiply two negatives and you get a positive
739779
ax²+bx+c=0\nDivide by a:\nx² + bx/a + c/a = 0\nSubtract c/a:\nx² + bx/a = -c/a\nAdd (b/(2a))²:\nx² + bx/a + b²/(4a²) = -c/a + b²/(4a²)\nFactor and simplify:\n(x+b/(2a))² = (b²-4ac)/(4a²)\nTake the square root:\nx+b/(2a) = ±√(b²-4ac)/(2a)\nSubtract b/(2a):\nx = (-b±√(b²-4ac))/(2a)
767438
b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
920068
The standard form of a quadratic:\n\nax^2 + bx +c\n\nIn your case a =1, and c=1. Since there is no x term, b=0
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc