y = sqrt X + 3sqrt X is the same as y = 4 sqrt X
x=y
The number you are referring to is the square root of the second number. In mathematical terms, if you have a number "x" and it produces the square of another number "y" when multiplied by itself, then x is the square root of y. For example, if x * x = y, then x is the square root of y.
If x equals the square root of ...., then you already have solved for x
sqrt(y*y) = sqrt(y2) =± y
5
A square root is taking a number X and determining which two same numbers times each other, or Y*Y equal X. If X equals 9, the square root of it is 3; that is 3 (Y) multiplied by 3 (Y) is 9.
-2 and -3Check:(-2) + (-3) = -5(-2)(-3) = 6Thus -2 and -3 are not the required numbers. let's find them: x + y = -6xy = -5 y = -x -6x(-x - 6) = -5-x^2 - 6x = -5x^2 + 6x = 5x^2 + 6x + 9 = 5 + 9(x + 3)^2 = 14x + 3 = (+ & -)square root of 14x = -3 (+ & -)square root of 14x = -3 + square root of 14 or x = - 3 - square root of 14y = -x - 6y = 3 - square root of 14 - 6 or y = 3 + square root of 14 - 6y = -3 - square root of 14 or y = -3 + square root of 14Check:(-3 + square root of 14) + (-3 - square root of 14) = -6(-3 + square root of 14)(-3 - square root of 14) = -5 ?(-3)^2 - (square root of 14)^2 = -5 ?9 - 14 = -5Check also tow other numbers.
y = sqrt X + 3sqrt X is the same as y = 4 sqrt X
3x y^3
x=y
x + y = 1xy = 1y = 1 - xx(1 - x) = 1x - x^2 = 1-x^2 + x - 1 = 0 or multiplying all terms by -1;(-x^2)(-1) + (x)(-1) - (1)(-1) = 0x^2 - x + 1 = 0The roots are complex numbers. Use the quadratic formula and find them:a = 1, b = -1, and c = 1x = [-b + square root of (b^2 - (4)(a)(c)]/2a orx = [-b - square root of (b^2 - (4)(a)(c)]/2aSox = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)x = [1 + square root of (1 - 4]/2x = [1 + square root of (- 3)]/2 orx = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;x = [1 + square root of (i^2 )(3)]/2x = [1 + (square root of 3)i]/2x = 1/2 + [i(square root of 3]/2 andx = 1/2 - [i(square root of 3)]/2Since we have two values for x, we will find also two values for yy = 1 - xy = 1 - [1/2 + (i(square root of 3))/2]y = 1 - 1/2 - [i(square root of 3)]/2y = 1/2 - [i(square root of 3)]/2 andy = 1 - [1/2 - (i(square root of 3))/2)]y = 1 - 1/2 + [i(square root of 3))/2]y = 1/2 + [i(square root of 3)]/2Thus, these numbers are:1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/22. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2Let's check this:x + y = 11/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1xy = 1[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:= 1/4 - [(3)(-1)]/4= 1/4 + 3/4= 4/4=1In the same way we check and two other values of x and y.
log of the square root of 'y' = 1/2 sqrt(y)
One step you usually want to take is to move any perfect square out of the radical. Example 1, with numbers: root(12) = root(4x3) = root(4) x root(3) = 2 root(3) Example 2, with variables: root(y cubed) = root(y squared times y) = root(y squared) times root(y) = y root(y)
y6
5x times the square root of y
If x is a multiplication symbol, then square root of 36(4)y^2 = square root of(4^2)(2^2)(y^2) = (4)(2)(y) = 8y If it is a variable x, then square root of 36x4y^2 = square root of (4^2)(x)(2^2)(y^2) = (4)(2)(y)(square root of x)= 8y(square root of x)