Need help solving for X in parabola equation.?
411391
y=(4b/a^2)x^2\n 1. sqaure root both sides to ge sqrt(y)=sqrt((4b/a^2)x^2)\n 2. the 2 in the numerator 2^2=4, the a in the denominator and the x comes out from under the root leaving the b under root so we have sqrt(y)=(2/a)*x*sqrt(b)\n 3. multiply both sides by (a/(2*aqrt(b)) to solve for x\n 4. x= + or -(a/(2*sqrt(b))*sqrt(y) two cases !!\n 5. plot the + and - solution seperately to obtain a parabola rotated 90 degrees clockwise
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
1057389
Look at the coefficient of the x^2 term.  If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum.  You have 2x^2, so you will have a minimum value.\n\nNext find the vertex.  The x-coordinate of the vertex lies at -b/2a.  So -8/(2)(2) = -8/4 = -2.  Plug this value in to get the y-coordinate of the vertex.  2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.\n\nSo you have a minimum value of -3.
961595
umm..are you missing an equals sign somewhere or am i really dumb? how do you figure out what x is unless you have both sides of the equation?
753539
Since f(x) is a parabola that opens upward, you need the minimum value, which is the y-coordinate of the vertex.  The x-coordinate is -(-12)/(2•3) = 2.  So the minimum y is at 3(2)² - 12(2) + 13 = 1, and your range is y >= 1.\n\nFor the second part, just solve 3x² - 12x + 13 = 28.\n\n3x² - 12x - 15 = 0\nx² - 4x - 5 = 0\n(x-5)(x+1) = 0\nx = 5 or x = -1.
867879
Just plug 0 in the function whenever you see y.\n\n4x + 3(0) + 24 = 0\n4x + 0 + 24 = 0  \n\nThen solve for x.\n\n4x = -24\nx = -6
798499
Given ax² + bx + c = 0, the quadratic formula tells us that:\n\nx = [ -b ± √(b²-4ac) ] ÷ 2a\n\nIn your problem, a=3, b=12, c=1  --  simply plug these values into the quadratic formula to solve for x.
715515
Profit- P(x)\n\nP(x) = R(x) - C(x)\nP(X)= 20x - .001x^2 - 2500 - 10x\nP(x)= 10x - .001x^2 - 2500\n\n\nWe have our equation, let's differentiate it (i'll use the shortcut and the power rule to speed things up).\n\nP'(x) = 10 - .002x\n\nThe zeroes of this will be local extrema of the original equation.  Since there is a negative sign in front of the .001x^2, that flips the parabola upside down and it will be the maximum point.\n\n0 = 10-.002x\nx=5000\n\nPlug this into the cost equation:  2500 + 10(5000) = 2500 + 50000 = 52500.\n\nHope this helps.
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
873365
I don't see any x  in the equations but I think the problem is that xmax< = xmin or Y max<=Y min  as a result the calulator could not calulate it that way try making Xmin < Xmax and Ymin < Ymax  this should solve the problem.
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
760976
The parabola y = (x+3)(x-5) crosses the x axis at -3 and 5.  It opens upward, so its vertex must be below the x axis.  So y < 0 for x in (-3, 5).
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
763084
or solve equation for x and y components separately. \nfor x    0 +3+?=0 \nfor y    3+0 +?=0