What are the roots of the equation x^2-5x-2 = 0?

Answer 1

ax² + bx + c = x² - 5x - 2 = 0
so the quadratic formula gives
x = [-b ± √(b² - 4ac)]/(2a)
= [-(-5) ± √((-5)² - 4*1*(-2))]/(2*1)
= [5 ± √33] /2

.,,

Answer 2

Completing the square

x² - 5x - 2 = 0

Transpose 2

x² - 5x - 2 + 2 = 0 + 2

x² - 5x = 2

x² - 5x +_____ = 2 +_____

x² - 5x + 25/4 = 2 + 25/4

(x - 5/2)(x - 5/2) = 8/4 + 25/4

(x - 5/2)² = 33/4

(√x - 5/2)² = ± √33 / √4

x - 5/2 = ± √33 / 2

Transpose 5/2

x - 5/2 + 5/2 = 5/2 ± √33/2

x = 5/2 ± √33/2

x = 5/2 ± 5.744562647 / 2

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Solving for +

x = 5/2 + 5/744562647 / 2

x = 10.74456265 / 2

x = 5.372281323. . .←. .Roots


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Solving for -

x = 5/2 - 5.744562647 / 2

x = - 0.744562647 / 2

x = - 0.372281323. . . ←. .Roots

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Answer 3

Using the standard form of solution for quadratic equations in the form of Ax^2+By+C=0...
x=(-B+/-sqrt(B^2-4AC))/2A
so
x=(5+/-sqrt(-5^2-4*1*-2))/2*1
then
x=(5+/-sqrt(25+8))/2
then
x=(5+/-5.74)/2
yielding the solutions
x= 5.37 and -0.37

happy holidays and happy algebra

Answer 4

Use the quadratic formula.

It comes out to (5+/-radical33)/2.