When using the quadratic formula to solve a quadratic equation (ax^2+bx+c=0), the discriminant is b-4ac .?
970904
discriminant is not b-4ac but b^2-4ac\nasuming a >0, if a < 0 the parabola is upside down\nthe above equation is equation of a parabola\n\nif b^2-4ac is  >0 this has 2 solutions so x axis or y=0 cuts the curve at 2 points\n\nif b^2 - 4ac = 0 then y axis is tangent to the curve\n\nif b^2-4ac < 0 then the curve is above x axis and y axis does not cut the curve
925102
Discriminant is part of the solution.  Just considering it tells you the types of solutions that you have.  \n\nThere are three possibilities:\n1. 2 real roots,\n2. 2 complex roots,\n3. 1 repeated real root.
686655
No prob dude... take a deep breath, grab a cold drink, and concentrate on this... ready? ok....\n\nOften, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. So, while factoring may not always be successful, the Quadratic Formula can always find the solution. \nThe Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:\n\n For ax2 + bx + c = 0, the value of x is given by:\n   \nNote that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
287408
y = 0\n4x^2 + 9 = 0\nget the discriminant\nb^2 - 4ac\n=0^2 - 4(4)(9)\n= -144\n\nTherefore, it has no real  x-intercepts but generally has 2 irreal (imaginary) x-intercepts^_^
684785
Presuming you are talking about a parabola,\n\nthe vertex is the maximum or minimum points of the parabola, the point at which the parabola curves back towards its original direction. For any parabola, the vertex's x-coordinate is the average of the two solution's coordinates.\n\nthe solution points, know as the root of the parabola, is the point at which the parabola intersects the x axis.\n\nfor the quadratic formula y=ax^2+bx+c,\n\nthe formula b^2-4ac is used to determine the amout of solutions a quadratic equation has. it is also called the discriminant\n\nif the answer is > 0, there are two solutions, or two places where the parabola intersects the x axis, if the answer is < 0, the parabola has NO solutions, it does NOT intersect the x axis, and if the answer is equal to 0, the parabola is tangent, or intersects the x axis at one point, and that point will be the VERTEX.\n\nhoped this helped ^^\n\nand if the
208415
D = (-6)^2 - 4*(-2)*(-3)= 36 - 24 = 12\nThe discriminant is 12, so it's positive, the roots are real and conjugate - it has 2 solutions.
626693
ax - b = -c\nax = b - c\nx = (b - c) / a.
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.
905417
By dividing the equation by 2:\n\nx^2 + 3x +1 = 0\n\nThe quadratic formula is = [-b +/- sqrt(b^2 - 4*a*c)]/2a\n\nWhere a =1;     b = 3;       c = 1.\n\nNow, we plug in this values in the quadratic formula, and we'll obtain a value of x, using the minus sign, and other, using the plus sign, as follows:\n\nx1= [ - 3 + sqrt( 3^2  - 4*1*1)]/(2*1)\nx1= [ - 3 + sqrt(9 - 4)]/2\nx1= [ - 3 + sqrt(5)]/2\nx1= [ -3 + 2.236]/2\nx1= -0.382\n\nx2= [ - 3 - sqrt( 3^2  - 4*1*1)]/(2*1)\nx2= [ - 3 - sqrt(9 - 4)]/2\nx2= [ - 3 - sqrt(5)]/2\nx2= [ -3 - 2.236]/2\nx2= -2.618
1072562
I'm reading the equation as \n\n4x^2 + 8x -1 = 0\n\nUsing quadratic formula x = (-8 +/- sqrt(64+16))/8 \n\nso x = -18/8 or 2/8 = -9/4 or 1/4\n\nThose are the roots.\n\nSubstitute each of those values as y into the second formula.
85829
I used the quadratic formula x={-b+-sqrootof b^2-4ac}/2a in order to find the factors on both the numerator and denominator\n\nthat leaves {(2x-1)(x-3)}/{4(x-3)(x-3)}\n\nClear (x-3) from both sides, that leaves\n\n(2x-1)/(4(x-3)) your answer
234536
Okay, equations of the form ax^2+bx+c=0 have the discriminant b^2-4ac. In your case a=4,b=-3,c=5. So you have to take (-3)^2-4*4*5=-71. Since the discriminant is negative, you have 0 real solutions. If it were 0, you would have 1 real solution, and if it were positive, you would have 2 real solutions. Hope that helps!
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)
516234
The discriminant = b^2-4ac = 1 - 4.2.(-4) = 33 > 0 \n⇒ We can get two unequal, real but irrational roots.\n\nThis is so, because, by actually calculating the roots of the equation applying formula. we find that:\nx= [1±√33]/4
207832
ax^2 + bx + c = 0\n\nx = [-b +/- sqrt(b^2 - 4ac)]/ 2a\ntwo answers since there is a +/- in there, and if the determinant (b^2 - 4ac) is negative, then the answers are complex conjugate pairs.\n\n1) x = 2 +/- sqrt(4-3) = 1,3 [real distinct]\n2) x = (-3 +/- sqrt(9 - 40))/4\n= -0.75 +/- i * sqrt(31/16)\n= -0.75 +/- i * 1.39 [complex conjugate pair]
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.