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Q: The square root of x-3 plus 5 equals x?

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The square root of x3 is x3/2 or x1.5. Either way it means sqrt(x3).

If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator

x3.

IMPOSSIBLE

No.

It is x = -5

No, it is not.

y2=x3+3x2

sqrt(x3 + x2) = x*sqrt(x+1) and that cannot be simplified further.

cube root of 216 = 6 > X= 6

(xn+2-1)/(x2-1)

square root of 12 = sq rt of (4 x3) = 2 x sq rt 3

9

The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1

x3 + 2x2 - 8x + 5 = 0 x(2x - 8) + 5 = 0

x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4

Realising that 125 is a perfect cube, this is actually pretty easy: x3 + 125 = 0 x3 = -125 x = -5

x3/x1/2 = x5/2.

X3-5=3 X3=3+5 X3=8 cube root of X3=cube root of 8 X=1.68 I hope this could help. There's not really any good ways to write algebra in a normal text box.

This integral is a bit complicated to try and type here, so I've included a link to a better representation of it below, under "related links". Also, I assume you do not mean to include the "plus 2" in the square root, as the integral becomes considerably more complicated then.

Square root is the same as raise to the 1/2 power, so multiply the exponents {(ab)c = abc}, so sqrt(x^3) = (x3)1/2 = x3/2

x3 - 36 has no real solution. It factors to(x - 3.30193)(x + (1.65096 - 2.85955i))(x + (1.65096 + 2.85955i)) where i is the square root of -1.

It is a polynomial if the square root is in a coefficient but not if it is applied to the variable. A polynomial can have only integer powers of the variable. Thus: sqrt(2)*x3 + 4*x + 3 is a polynomial expression but 2*x3 + 4*sqrt(x) + 3 is not.

2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7

Integral of x3/2dx using power rule = (5/2)x5/2 2.5 times the square root of x to the fith.