show that x=2 is one root of this ...?
817379
y=(x^2-3x)^2\ny=x^4-9x^2\nAnd then I'm confused.
797453
you set the bottom of the equation equal to 0. So x^2 - x = 0, which leads to two cases. The first one is x = 0. The second one is x-1 = 0, or x = 1.
832245
square root by x both sides\n\nso\n\n5 - 3 = X Root 2\n2= xroot 2\nso x = 1.
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
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}
338782
sqrt x^2 = x is  not an identity.Its true only 4 positive values of x.\nThe reason is explained as follows:-\n\nConsider x > 0\nthen x^2 > 0 (of course,yes)\nthen sqrt x^2 = ± x.\nsince x > 0(as considered above),\nhence sqrt x^2 = x (for all x > 0).\n\nNow,Consider x < 0\nthen x^2 > 0(of course,yes)\nthen sqrt x^2 = ± x.\nsince x < 0(as considered above),\nhence sqrt x^2 = - x (for all x < 0).\n\nSo in this case,after taking sqrt,we have 2 insert a negative sign 2 balance the equation.\n\nHence,sqrt x^2 = x is not an identity(coz its true only 4 positive values of x).
837906
x is (2n+1)th root of 1\n\nso x = e^(kpi/(2n+1)) k is from 1... 2n\n\n1+x+x^2+....+x^2n = 0\n\nx+x^2+.... + x^2n = -1\nthe real part of this is \n\ncos(2pi/(n+1))+cos(4pi/(n+1)+ ... = -1
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
1149923
(2,-4) if f(x) is defined for x=2
220918
2*x^2 - 3*x - 2 = 0 <=> x = -1/2 V x = 2 \n\n==> 2*x^2 - 3*x - 2 <=0 <=> x = [-1/2 , 2]
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
1056888
2(x^2-4x)-1=\n2(x^2-4x+4) -8-1=\n2(x-2)^2-9
809248
Just break it into parts. X=1x. X can be any other variable but the thing that matters is that all the variables are the same. 1x-.8x=.2x. Then .2x-2.9x= -2.7x. It simplifies down to -2.7x but there is know way to figure out what the value of X is. If the equation was x-.8x-2.9x=any thing a value for x would be possible to find but there is known and so there is no way to find the value of x on it's own.
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
750535
if f(x) change of sign twice\n\nso number of positive roots 0 or 2 based on Descartes rule\n\nfrom the given case 2 positive so 1 negative
206555
is the question x=x^2?\nif so 1 is the only number that can always satisfy that
762440
Just split it into parts - when is x-x^2<0, and when is it >0?\nThat bit has roots 0 and 1, so it is positive between 0 and 1, negative everywhere else.\n\nThus, ∫(x=-1 to x=0) (x^2-x) dx + ∫(x=0 to x=1) (x-x^2) dx + ∫(x=1 to x=3) (x^2 - x) dx\n\n= [(1/3 x^3 - 1/2 x^2) from -1 to 0] + [(1/2 x^2 - 1/3 x^3) from 0 to 1] + [(1/3 x^3 - 1/2 x^2) from 1 to 3]\n= (0- -5/6) + (1/6 - 0) + (9/2 - -1/6)\n= 17/3.