If one root of x squared + 2x - c = 0 is x = 1, what is the value of c?
540886
According to the question the answer is 3
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
602488
1-x = -(x-1)\nx^3 - 1 = (x-1)(x^2+x+1)\nAnswer: -(x-1)(x^2+x+1)
195280
In this polynomial: a=6, b=-2, c=-20.\nSub a, b, and c into this equation: \nx= [-b+ square root the whole thing that inside this bracket(b*b -4ac)]/2a\n\nx'= [-b - square root the whole thing that inside this bracket(b*b -4ac)]/2a
429058
x==(-b+/-sqrt(b^2-4ac))/(2a).
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
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".
834259
sq root of X-1 cannot come out to be negative...\nso...x-1>0 or x-1 = 0\n       x>1     or x=1\nDomain: {x|xer, x > or = 1 }\nRange: {y|yer, y> or = 0}
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
855361
2x^2 - 9x - 5=0\n(2x+1)(x-5)=0\n2x+1=0\nx=-1/2\nx-5=0\nx=5\n\nroots are \nx=5, -1/2
282131
Note that although x1=x2 satisfy the equations but some distinct values for x1 and x2 can satisfy it too: \n\nsuppose x1 and x2 are distinct and so x1-x2 is not zero\nthen eq1 is:\n\n4(x1-x2)(x1+x2)-4(x1-x2)=0\n\n---> 4(x1+x2)-4=0\n---> x1+x2=1\n\nand eq2 is:\n\n-3(x1-x2)(x1+x2)(x1^2+x2^2)+2(x1-x2)(x1+x2)=0\n\nomit (x1-x2)\nand substitute x1 by 1-x2 to get:\n(1-x2)^2+x2^2=2/3\n\n---> x2=(3+-sqrt(3))/6\n\nso we have two values for x1 and x2 which satisfy the equation and x1 and x2 are distinct.
862692
y" + y = 1\n\n1=sin(x) + xcos(x)\n\nEquation should read 1=sin^2(x) + xcos^2(x)\nand x value is elements of real numbers.
572493
=x(x^3-2)\n--> x=0 or x=2^(1/3) or -2^(1/3)
268383
2x^2+6x-1x-3\n2x(1x+3)-1(1x+3)=0\n(2x-1)(x+3)=0\nx=1/2 or -3
936579
your question seems to be incomplete.\nso far....\nwidth=x\nlength= 2x-1
345141
0 solutions!\n\nMove sqrt(x+8) on the other side and then raise by the power of 2. You will get\n\nx-4=4+4*sqrt(x+8)+x+8    (btw if there is a solution it has to be greater or equal to 4)\n\nthis is equivallent to \n\nsqrt(x+8)=-4\n\nWhich is impossible by the definition of sqrt(). It always has to be greater or equal to 0.
230663
you need imaginary numbers to solve this one. Use the quadratic formula to solve ax^2 + bx + c = 0:\n\nx = -b +- sqrt(4ac)\n\nin this case:\na=1, b=2, c=-1\n\nso x has two solutions:\nx=-2+i  or x= -2-i \nwhere i=sqrt of -1 (the unit imaginary number)