Answer:
[tex]\large\boxed{x=\dfrac{13-\sqrt{265}}{2}\ and\ y=\dfrac{13+\sqrt{265}}{2}}\\\\.\qquad\qquad\qquad\qquad or\\\\\boxed{x=\dfrac{13+\sqrt{265}}{2}\ and\ y=\dfrac{13-\sqrt{265}}{2}}[/tex]
Step-by-step explanation:
[tex]x,\ y-two\ numbers\\\\\text{We have the equations}\\\\x+y=13\ \text{and}\ xy=-24\\\\x+y=13\qquad\text{subtract y from both sides}\\x=13-y\qquad\text{substitute it to the second equation}\\\\(13-y)y=-24\qquad\text{use distributive property}\\13y-y^2=-24\qquad\text{add 24 to both sides}\\-y^2+13y+24=0\qquad\text{change the signs}\\y^2-13y-24=0[/tex]
Use the quadratic formula:
[tex]ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a},\ x_2=\dfrac{-b+\sqrt\Delta}{2a}[/tex]
[tex]y^2-13y-24=0\\\\a=1,\ b=-13,\ c=-24\\\\\Delta=(-13)^2-4(1)(-24)=169+96=265\\\\y_1=\dfrac{-(-13)-\sqrt{265}}{2(1)}=\dfrac{13-\sqrt{265}}{2}\\\\y_2=\dfrac{-(-13)+\sqrt{265}}{2(1)}=\dfrac{13+\sqrt{265}}{2}[/tex]
Put the values of y to the equation x = 13 - y:
[tex]x_1=13-\dfrac{13-\sqrt{265}}{2}=\dfrac{26}{2}-\dfrac{13-\sqrt{256}}{2}=\dfrac{26-13-(-\sqrt{256})}{2}=\dfrac{13+\sqrt{265}}{2}\\\\x_2=13-\dfrac{13+\sqrt{265}}{2}=\dfrac{26}{2}-\dfrac{13+\sqrt{265}}{2}=\dfrac{26-13-\sqrt{256}}{2}=\dfrac{13-\sqrt{265}}{2}[/tex]
