Methods of solving quadratics equations..?

Question

Methods of Solving Quadratic Equations:
Apply the square root Property of Equations to find the solution set.

x^2 = 9

{___,____}

(3x-1)^2 = 5

(__+ sqrt ___/__, ___- sqrt__/__)

Help! Thanks!

Answer 1

x^2 = 9

{_3__,_-3___}

(3x-1)^2 = 5

(_1_+ sqrt _5__/_3_, _1__- sqrt_5_/_3_)

Answer 2

First check out the equation in common style. If a factoring is clean, use it. If factoring isn't glaring, it is going to nonetheless be factorable, basically you do no longer see it. if so compute the discriminant b² – 4ac. you at the instant are not dropping time, through fact if no longer factorable, you will desire this element besides. If the discriminant is a suited sq., through fact of this there are rational roots. At this ingredient you may take yet another crack at factoring with self belief that there is an answer, or merely proceed and use the quadratic formula. Exception: If equation is lacking the x-time era, use the sqaure root technique. it is is you have ax² – c = 0. It hardly ever will pay to end the sqaure, becasue the quadratric formula is the genral answer once you clean up via polishing off the sqaure, so which you would be able to besides use the formula. polishing off the sq. is used someplace else for different techniques, yet provides little right here. Al that bneing pronounced, there are m any techniques to help in factoring. commonly of thumb, if there are too many opportunities in facrtoring or the numbers are large, then you certainly would as welll bypass to the quadratic formula. once you may element, it is continuously swifter, so continuously provide it a attempt first, then fall lower back to the discriminant/quadratic formula. wish this facilitates. Factoring is slightly an artwork. journey will make you extra proficient. while attempting to element, fill interior the indicators first and then attempt interior the numbers.

Answer 3

sqrt is an operator just like any other (e.g., +, - , / , X); so keep an equation balanced, you have to use the operator on both sides of the equation.

Thus:

sqrt(x^2) = sqrt(9)
x = +- 3

sqrt(3x - 1)^2 = sqrt(5)
3x - 1 = sqrt(5)
3x = sqrt(5) + 1
x = [sqrt(5) + 1]/3

Answer 4

X^2 = 9
X = (3, -3)

(3x-1)^2 = 5
(3x-1)(3x-1) = 5
9x^2 - 3x - 3x +1 = 5
9x^2 - 6x - 4 = 0
6+/-sqrt([-6]^2 - 4 [9][-4])/2(9)
6 +/- sqrt(36 + 144)/18
6 +/- sqrt(180)/18
6 +/- 13.4164/18
19.4164/18 & -7.4164/18
1.0786 & -0.412022

Answer 5

take the Square root of both sides

Answer 6

first one's root 9, so +/- 3

second one is:
3x-1=root5
x=(root5+1)/3