Question

What are the exact solutions of x2 - 3x - 1 = 0?

Answer 1

$x^2-3x-1=0\\a=1;\ b=-3;\ c=-1\\\\\Delta=b^2-4ac\\\Delta=(-3)^2-4\cdot1\cdot(-1)=9+4=13 \ \textgreater \ 0\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{13}\\\\x_1=\dfrac{-(-3)-\sqrt{13}}{2\cdot1}=\dfrac{3-\sqrt{13}}{2}\\\\x_2=\dfrac{-(-3)+\sqrt{13}}{2\cdot1}=\dfrac{3+\sqrt{13}}{2}$

Answer 2

Answer:

$x_{1}=\frac{-3+\sqrt{13} }{2}$

$x_{2}=\frac{-3-\sqrt{13} }{2}$

Step-by-step explanation:

the equation is: $x^2-3x-1=0$

you have a quadratic equation of the form:

$ax^2+bx+c=0$

where $a=1,b=-3,c=-1$

this can be solved using the general formula:

$x=\frac{-b+-\sqrt{b^2-4ac} }{2a}$

substituting all the known values:

$x=\frac{-(-3)+-\sqrt{(-3)^2-4(1)(-1)} }{2(1)} \\x=\frac{-3+-\sqrt{9+4} }{2}\\ x=\frac{-3+-\sqrt{13} }{2}$

one value for x will be found using the plus sign before the square root:

$x_{1}=\frac{-3+\sqrt{13} }{2}$

and the other will be found using the negative sign before the square root:

$x_{2}=\frac{-3-\sqrt{13} }{2}$