Respuesta :
Given:
The equation are
[tex]y=3x+2[/tex]
[tex]y=2x^2+7x-11[/tex]
To find:
The coordinates of intersection points.
Solution:
We have,
[tex]y=3x+2[/tex] ...(i)
[tex]y=2x^2+7x-11[/tex] ...(ii)
From (i) and (ii), we get
[tex]3x+2=2x^2+7x-11[/tex]
[tex]0=2x^2+7x-11-3x-2[/tex]
[tex]0=2x^2+4x-13[/tex]
If a quadratic equation is [tex]ax^2+bx+c=0[/tex], then by quadratic formula:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
In the equation [tex]0=2x^2+4x-13[/tex], we have [tex]a=2,b=4,c=-13[/tex].
Using quadratic formula, we get
[tex]x=\dfrac{-4\pm \sqrt{4^2-4(2)(-13)}}{2(2)}[/tex]
[tex]x=\dfrac{-4\pm \sqrt{16+104}}{4}[/tex]
[tex]x=\dfrac{-4\pm \sqrt{120}}{4}[/tex]
Now,
[tex]x=\dfrac{-4+\sqrt{120}}{4}[/tex] and [tex]x=\dfrac{-4-\sqrt{120}}{4}[/tex]
[tex]x=1.7386[/tex] and [tex]x=-3.7386[/tex]
[tex]x\approx 1.74[/tex] and [tex]x\approx -3.74[/tex]
Putting x=1.74 in (i), we get
[tex]y=3(1.74)+2[/tex]
[tex]y=5.22+2[/tex]
[tex]y=7.22[/tex]
Putting x=-3.74 in (i), we get
[tex]y=3(-3.74)+2[/tex]
[tex]y=-11.22+2[/tex]
[tex]y=-9.22[/tex]
Therefore, the coordinates of the intersection points are (1.74, 7.22) and (-3.74,-9.22).
