Answer:
Part 1) The exact solutions are
[tex](\frac{-1+\sqrt{21}} {2},4+\sqrt{21})[/tex] and [tex](\frac{-1-\sqrt{21}} {2},4-\sqrt{21})[/tex]
Part 2) (1.79, 8.58)
Step-by-step explanation:
we have
[tex]y=x^{2} +3x[/tex] ----> equation A
[tex]y=2x+5[/tex] ----> equation B
we know that
When solving the system of equations by graphing, the solution of the system is the intersection points both graphs
Find the exact solutions of the system
equate equation A and equation B
[tex]x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +x-5=0[/tex]
so
[tex]a=1\\b=1\\c=-5[/tex]
substitute in the formula
[tex]x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}[/tex]
[tex]x=\frac{-1\pm\sqrt{21}} {2}[/tex]
so
The solutions are
[tex]x_1=\frac{-1+\sqrt{21}} {2}[/tex]
[tex]x_2=\frac{-1-\sqrt{21}} {2}[/tex]
Find the values of y
First solution
For [tex]x_1=\frac{-1+\sqrt{21}} {2}[/tex]
[tex]y=2(\frac{-1+\sqrt{21}} {2})+5[/tex]
[tex]y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}[/tex]
The first solution is the point [tex](\frac{-1+\sqrt{21}} {2},4+\sqrt{21})[/tex]
Second solution
For [tex]x_2=\frac{-1-\sqrt{21}} {2}[/tex]
[tex]y=2(\frac{-1-\sqrt{21}} {2})+5[/tex]
[tex]y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}[/tex]
The second solution is the point [tex](\frac{-1-\sqrt{21}} {2},4-\sqrt{21})[/tex]
Round to the nearest hundredth
First solution
[tex](\frac{-1+\sqrt{21}} {2},4+\sqrt{21})[/tex] -----> [tex](1.79,8.58)[/tex]
[tex](\frac{-1-\sqrt{21}} {2},4-\sqrt{21})[/tex] -----> [tex](-2.79,-0.58)[/tex]
see the attached figure to better understand the problem
