Help solving Algebra 2 questions?

Question

I have to re-take an Algebra 2 test, but I don't understand how to solve a couple of questions that I got wrong. If anyone can explain to me how to solve them, help me out :) Thanks!!!

P.s.- you have to solve and check answers for extraneous roots)

1. √x - 3 = x - 5

2. √14x - 5 = x + 2

3. √2x - 3 =  √3x - 2 

4. 3√2x - 3 = 7         (The "3" before the square root is actually an exponent.)

Thanks so much if you can help me out.

LS · Accepted Answer

1. To isolate the root, make the equation:
√x = x - 2
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 
 √1 -3 = 1 - 5 -2 =/= -4 so 1 is extraneous

2. √14x = x + 7 
14x = x^2 + 14x +49 
 x^2 + 49 = 0
 no solution 

3. 
-3 = √x - 2
 √x = -1
 x = -1 
 no solution  since we can't solve √of a neg. number

4.
3√2x = 10 
 2x = 10^3 (1000) --> x = 500 
Check and you get 7 = 7

? · Answer

ending up the sq. 2x² + 20x = -38 x² + 10x = - 19 [ Divide by using common element of two] x² + 10x + 25 = -19 + 25 [upload 25 to the two the factors so as that L.H.S takes this sort 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 on condition that the (?6)² and (-?6)² are the two reminiscent of 6] hence 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 region 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 !!!

Trending News

Help solving Algebra 2 questions?

2 Answers