How do you solve this?

Question

x^2 - square root of 5x - 1 = 0

again...sorry for the poor notation

I got square root of 5 + OR - 3/2 Let me know if it's correct

x^2-√5x -1=0


5x is under the √

Answer 1

Do you mean:
x^2 - SQRT(5)*x - 1 = 0

x = [SQRT(5) +/- SQRT( 5 - 4*1*-1)]/2
x = 0.5 * [SQRT(5) +/- SQRT(9)]
x = 0.5*[SQRT(5) +/- 3]

(The SQRT(5) is also over 2)


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Or do you mean:
x^2 - SQRT(5x) -1 = 0
which is a more difficult problem (solved as a 4th degree polynomial)

SQRT(5x) = x^2 - 1
5x = (x^2 - 1)^2 = x^4 -2x^2 +1

x^4 -2x^2 -5x +1 = 0

Answer 2

x²-√5x -1=0
delta = (-√5)² - 4*1*-1
delta = 5 + 4
delta = 9

x = [-(-√5) +/- √9] : 2*1
x' = [√5 + 3] : 2 = √2,5 + 1,5 = 3,08 round
x" = √2,5 - 1,5 = 1,58 - 1,5 = 0,08 round
solution: {x belongs to R| x = 3,08 or x = 0,08}
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Answer 3

I got 1/4 but i dunno whether i have interpret your qn wrongly.

x^2-(5x-1)^1/2=0

x^2 = (5x-1)^1/2

then get rid of the square on the LHS by square rooting the RHS

x = 5x-1

4x = 1

x = 1/4

Answer 4

You did alright IF you meant that you got (-sqrt(5) (+/-) 3)/2
Regards
Tonio

Answer 5

sqrt (5) or sqrt (5x) ????????????

Answer 6

wut do u mean, where is the problem exactly