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[SOLVED] Intersection Points & Finding Unknown Variable

confusedatmath

New member
Jan 2, 2014
14
The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = x^2 + x − 2 in two distinct points if

I first made the equations equal each other

x + k = x^2 + x − 2
0 = x^2 -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2
 

mente oscura

Well-known member
Nov 29, 2013
172
The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = + x − 2 in two distinct points if

I first made the equations equal each other

x + k = + x − 2
0 = -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2
Hello.

Check the wording of the question. The parable, need that 'x' is not high to the power 1. (would be a straight line)

Regards.
 

confusedatmath

New member
Jan 2, 2014
14
Hello.

Check the wording of the question. The parable, need that 'x' is not high to the power 1. (would be a straight line)

Regards.

i fixed it :p read again, it was a mistake i forgot the x^2
 

mente oscura

Well-known member
Nov 29, 2013
172
The line with equation y = x + k, where k is a real number, intersects the parabola with equation y = x^2 + x − 2 in two distinct points if

I first made the equations equal each other

x + k = x^2 + x − 2
0 = x^2 -2 -k

From here i thought you use the discriminate a=1 b=o c=-2-k

but this isn't right, because the answer to choose from

k < − 2

k > − 2

k = − 2

k < 2

k ≠ 2
Now yes.

[tex]0=x^2-2-k[/tex]

[tex]x^2=k+2[/tex]

[tex]x=\pm{} \sqrt{k+2}[/tex]

1º) [tex]k<-2 \rightarrow{}x \cancel{\in}{R}[/tex]

2º) [tex]k>-2 \rightarrow{}x \in{R}[/tex]

3º) [tex]k=-2 \rightarrow{}x=0[/tex], only a breakpoint.

Regards.