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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.

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6 votes

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).

User Swihart
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