Help solving Algebra 2 questions?

In order to solve for a root, you must first isolate the root.

1. To isolate the root, make the equation:

√x = x - 2; then you need to get rid of the radical sign so square each side (the right is a polynomial)

x = x^2 - 4x + 4 ---> x^2 - 5x + 4 = 0; factor x = 4 and 1 check to see if they work

√4 - 3 = 4 - 5 ---> -1 = -1 so 4 is a solution; √1 -3 = 1 - 5 ---> -2 =/= -4 so 1 is extraneous

2. Do the same:

√14x = x + 7 ---> 14x = x^2 + 14x +49 ---> x^2 + 49 = 0 ---> no solution

3. Here subtract the radicals:

-3 = √x - 2 ---> √x = -1 ---> x = -1 ---> no solution (can't solve √of a neg. number)

4. The 3 means a cubed root:

3√2x = 10 (you need to cube instead of square) ---> 2x = 10^3 (1000) --> x = 500

Check and you get 7 = 7

Hope this helped!

ending up the sq. 2x² + 20x = -38 x² + 10x = - 19 [ Divide by way of worry-unfastened component to two] x² + 10x + 25 = -19 + 25 [upload 25 to the two the climate so as that L.H.S takes the style of x² + 2xy + y²] (x+5)² = 6 [x² + 2xy + y² = (x+y)²] x + 5 = ± ?6 [Take sq. root of the two aspects , R.H.S ought to be ± ?6 for the explanation that (?6)² and (-?6)² are the two resembling 6] subsequently x= -5 + ?6 or x = -5 - ?6 Quadratic equation Roots of the equation 2x² + 20x = - 38 2x² + 20x + 38 = 0 [Taking -38 to the L.H.S] Roots are -(b ±?(b² - 4ac) )/ 2a the area a= 2 b = 20 and c = 38 = ( -20 ±?( (20)² - 4(2)(38) / 2(2) ) = ( -20 ±?( 4 hundred - 304) / 4) = ( ( -20 ±?ninety six / 4) = ( -20 ±?sixteen * ?6) / 4) = (-20 ±4?6)/4 = -5 ±?6 ie : -5 + ?6 and -5 - ?6 are the roots Cheers !!!

1.) 2.35

2.) 4/3

3.) 1.2

4.) 6/7