# For what range of k values will the linear-quadratic system below have 2 solutions? y=2x^2+4+5 y=kx-3

So you have `2x^2+4+5 = kx-3`

`2x^2-kx+12=0` this equation will have two solutions if and only if its discriminant `D = b^2 - 4ac neq 0` where `a, b, c` are coeffitients of quadratic equation.

`k^2 - 4 cdot 2 cdot 12 neq 0 implies k neq pm sqrt(96)`

So...

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So you have `2x^2+4+5 = kx-3`

`2x^2-kx+12=0` this equation will have two solutions if and only if its discriminant `D = b^2 - 4ac neq 0` where `a, b, c` are coeffitients of quadratic equation.

`k^2 - 4 cdot 2 cdot 12 neq 0 implies k neq pm sqrt(96)`

So for `k = pm sqrt(96)` you would have 1 solution and for all other values of `k` you would have 2 solutions which would be complex if discriminant is negative and real if discriminant is positive.

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We are given `y=2x^2+4x+5` The slope of the tangent line is given by `y'=4x+4` .

The system will have two solutions if the line intersects the parabola twice. It will do so between the values where the line is tangent to the parabola.

A point on the parabola has coordinates `(x,2x^2+4x+5)` . The slope between this point and the point (0,-3) (Which lies on any line y=kx-3) is:

`m=(2x^2+4x+5-(-3))/(x-0)=(2x^2+4x+8)/x` . If the linewith this slope is tangent to the curve its slope is also give by m=4x+4.

So `(2x^2+4x+8)/x=4x+4`

`2x^2+4x+8=4x^2+4x`

`2x^2=8==>x=+-2`

If x=2 then the point on the parabola is (2,21) and the slope between (2,21) and (0,-3) is 12, while the slope of the tangent line at x=2 is 4(2)+4=12

Also, if x=-2 the pointon the parabola is (-2,5) and the slope from (-2,5) to (0,-3) is -4, while the slope of the tangent line at x=-2 is 4(-2)+4=-4.

If x=2, then k=12 and if x=-2 then k=-4. So any value of k such that -4<k<12 will cause the line y=kx-3 to intersect the parabola twice.

The graphs of the parabola and the two tangent lines:

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