How do you do this problem x^2-4x-1=0?

Question

x^2= x to the second power, you are suppose to use the quadratic formula -4 + or - the square root of b^2-4ac/2a

how do you get 2+ or- square root of 5?

Answer 1

Sounes like you answered your own question: use the quadratic formula.

x = [ -b ± √(b^2 - 4ac) ] / 2a
x = [ 4 ± √(16 - 4(1)(-1)) ] / 2(1)
x = [ 4 ± √20] / 2
x = [ 4 ± 2√5] / 2
x = 2 ± √5

Answer 2

The quadratic formula always works for any quadratic equation (equations with a degree of 2). Because this equation cannot be solved by the simpler method of factoring, simply use the quad. formula:

x = -b (+ or -) the square root of (b^2 - 4ac), all divided by 2a.

Your equation is written in the form ax^2 + bx + c, so plug in the values for a, b, and c in the formula:

a = 1, b = -4, c = -1

x = -(-4) [+ or -] the square root of ((-4)^2 - 4(1)(1)), all over 2(1)
= [4 [+ or -] the square root of (16 - 4)]/2
= [4 [+ or -] 2[root]3]/2
= 2 [+ or -] [root]3
x = 2 + square root of 3, or 2 - square root of 3.

Answer 3

yes, you are supposed to use the quadratic formula to find the roots since you cannot factor that equation so easily.
x² - 4x - 1 = 0

quadratic formula: [-b ± √(b² - 4(a)(c))]/2a

x = [-(-4) ± √(16 - 4(1)(-1))]/2(1)
x = (4±√20)/2 = (4±2√5)/2 = 2±√5
so x ~= 4.236067977 or -0.2360679775

Answer 4

x^2-4x-1=0
comparing the expression with the standard quadratic equation ax^2+bx+c=0,we get
a=1,b= -4 and c= -1
Therefore
x={-b+-sqrt(b^2-4ac)}/2a
={4+-sqrt(16+4)}/2*1
=(4+-sqrt20)/2
=(4+-sqrt(4*5)/2
=(4+-2sqrt5)/2
=2+-sqrt5 ans

Answer 5

quadratics

ax^2 + bx + c

a= 1
b= -4
c= -1

the quadratic formula - x = [-b +/- sqrrt(b^2 - 4ac)] / 2a

replace the thingys to get [4 +/- sqrrt(-4^2 - 4*1*-1)] / 2*1

[4 +/- sqrrt(20)] / 2

(4 +/- 4.47) / 2

x = 4.24 , -0.24

yeah

Answer 6

x = [ 4 ± √(16 + 4) ] / 2
x = [ 4 ± √(20) ] / 2
x = [ 4 ± 2√(5) ] / 2
x = 2 ± √5