Question

The straight line y = 3x +2 intersects the curve y = 2x^2 + 7x-11 at two points. Find the coordinates of these two points Give your answers correct to 2 decimal places.

Answer 1

Given:

The equation are

`y=3x+2`

`y=2x^2+7x-11`

To find:

The coordinates of intersection points.

Solution:

We have,

`y=3x+2` ...(i)

`y=2x^2+7x-11` ...(ii)

From (i) and (ii), we get

`3x+2=2x^2+7x-11`

`0=2x^2+7x-11-3x-2`

`0=2x^2+4x-13`

If a quadratic equation is
`ax^2+bx+c=0`, then by quadratic formula:

`x=(-b\pm √(b^2-4ac))/(2a)`

In the equation
`0=2x^2+4x-13`, we have
`a=2,b=4,c=-13`.

Using quadratic formula, we get

`x=(-4\pm √(4^2-4(2)(-13)))/(2(2))`

`x=(-4\pm √(16+104))/(4)`

`x=(-4\pm √(120))/(4)`

Now,

`x=(-4+√(120))/(4)` and
`x=(-4-√(120))/(4)`

`x=1.7386` and
`x=-3.7386`

`x\approx 1.74` and
`x\approx -3.74`

Putting x=1.74 in (i), we get

`y=3(1.74)+2`

`y=5.22+2`

`y=7.22`

Putting x=-3.74 in (i), we get

`y=3(-3.74)+2`

`y=-11.22+2`

`y=-9.22`

Therefore, the coordinates of the intersection points are (1.74, 7.22) and (-3.74,-9.22).