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Q: Sqrt x-7-5 equals 0

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It is sqrt(0), which equals 0.

Two. x = + i sqrt(8.4) x = - i sqrt(8.4)

x = 2.5 + sqrt(29) and x = 2.5 - sqrt(29)

[ sqrt(8)/sqrt(2) ] + [ 10/5 Hertz-seconds ] + cos(180Â°) + Ïµ-0 - log(10) + [ sqrt(6) ]2

x2 - 2x - 5 = 0 By the quadratic equation, x = {2 +/- sqrt[(-2)2 - 4*1*(-5)]}/2 = {2 +/- sqrt(4+20)}/2 =1 +/- sqrt(6)

3p2 - 21p + 35 = 0 Then p = [21 +/- sqrt(212 - 4*3*35)]/(2*3) = [21 +/- sqrt(441-420)]/6 = [21 +/- sqrt(21)]/6 = 2.736 and 4.264

0

Distance between (3, 0) and (0, -6) = sqrt[(3 - 0)2 + (0 - -6)2] = sqrt(32 + 62) = sqrt(45) = 3*sqrt(5) or 6.71 approx.

x = [-6 +/- sqrt(62-4*1*41)]/2 =[-6 +/- sqrt(36-164)]/2 =[-6 +/- sqrt(-128)]/2 = -3 +/- sqrt(-32) = -3 +/- i*sqrt(32) where i = the imaginary sqrt of -1

No.sqrt(2) - sqrt(2) = 0, which is rational.No.sqrt(2) - sqrt(2) = 0, which is rational.No.sqrt(2) - sqrt(2) = 0, which is rational.No.sqrt(2) - sqrt(2) = 0, which is rational.

x=(6±sqrt(36-4*6))/2 x=3±(sqrt(12)/2) x=3±sqrt(3)

The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.

2x2 - x - 5 = 0 x = [1 +/- sqrt(1 + 40)]/4 = [1 +/- sqrt(41)]/4 = -1.35078 and 1.85078

2x2 - 8x + 3 = 0 needs to be solved using the quadratic formula.x = (4 + sqrt(10))/2 or x = (4 - sqrt(10))/2

x2+2x-4 = 0 x = (-2 plus/minus sqrt (4-(4x1x-4)) / 2 x = (-2 plus/minus sqrt 20)/2 x = (-2 - sqrt 20)/2 or (-2 + sqrt 20)/2 x = -3.236 or 1.236

Is this 3*sqrt((x+25)*x)=0 or 3*sqrt(x+25x)=0 or 3*sqrt(x)+25x=0 or 3*sqrt(x+25)*x=0? I'm not going to bother answering all possible interpretations, but here's a hint: square the equation to get rid of the square root operation first. Then solving it will be easy.

Distance = sqrt(82 + 172) = sqrt(64 + 289) = sqrt(353) = 18.79That would be points (0, 0) and (8, 17)Distance = sqrt[(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt[(17 - 0)2 + (8 - 0)2]Distance = sqrt(172 + 82)Distance = sqrt(289 + 64)Distance = sqrt(353)

2x2-5 = 0 +5 +5 2x2 = 5 /2 /2 x2 = 5/2 sqrt(x2) = sqrt (5/2) x = sqrt(5/2) Ps: Sqrt means square root

The roots of the equation are [5 +/- sqrt(11)]/2 = 4.158 and 0.842

2w2 + 22w + 7 = 0 w = [-22 +/- sqrt(22*22 - 4*2*7)] / (2*2) = [-22 +/- sqrt(484 - 56)] / 4 = [-22 +/- sqrt(428)] / 4 so w = -0.328 or w = -10.672

x2 + 4x - 3 = 0 The quadratic formula is x = {-4 Â± sqrt[(-4)2 - 4*1*(-3)]}/2 = {-4 Â± sqrt[16 + 12]}/2 = {-4 Â± sqrt[28]}/2 = {-4 Â± 2*sqrt[7]}/2 = -2 Â± sqrt[7]

sqrt(52)+x=0 Thus, x=-sqrt(52)

4n2-25 = 0 4n2=25 n2=6.25 n=sqrt(6.25) n= +2.5 or -2.5

x2 - x + 3 = 0x = [1 +/- sqrt(1-12)]/2 = [1 +/- i*sqrt(11)]/2 where i is the imaginary square root of -1.

You cannot prove that sqrt(3)/2 = 0 because it is simply not true! The solution to the equation is theta (or, tita as you like to call it) = pi/6c or 30 degrees. The cosine of that angle is sqrt(3)/2 but that is NOT the same as it being 0.

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