[ sqrt(8)/sqrt(2) ] + [ 10/5 Hertz-seconds ] + cos(180°) + ϵ-0 - log(10) + [ sqrt(6) ]2
2 k^2 - k - 4 = 0 2 (k^2 - (1/2)k - 2) = 0 2 ((k - 1/4)^2 - 1/16 - 2) = 0 2 ((k - 1/4)^2 - 33/16) = 0 2 (k - 1/4 - sqrt(33)/4)(k - 1/4 + sqrt(33)/4) = 0 32 (4k - 1 - sqrt(33))(4k - 1 + sqrt(33)) = 0
There are no real root. The complex roots are: [-5 +/- sqrt(-3)] / 2
Can be irrational or rational.1 [rational] * sqrt(2) [irrational] = sqrt(2) [irrational]0 [rational] * sqrt(2) [irrational] = 0 [rational]
(0, 4) and (- 4, 6) ???Distance = sqrt[(Y2 - Y1)2 + (X2 - X1 )2]Distance = sqrt[(6 - 4)2 + (- 4 - 0)2]Distance = sqrt( 4 + 16)Distance = sqrt(20)==============
It is sqrt(0), which equals 0.
Two. x = + i sqrt(8.4) x = - i sqrt(8.4)
[ sqrt(8)/sqrt(2) ] + [ 10/5 Hertz-seconds ] + cos(180°) + ϵ-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
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
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
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.
x=(6±sqrt(36-4*6))/2 x=3±(sqrt(12)/2) x=3±sqrt(3)
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
The two solutions are the conjugate complex numbers, +i*sqrt(11) and -i*sqrt(11), where i is the imaginary square root of -1.