# Simultaneous Linear and Quadratic Equations

• Mar 21st 2009, 11:59 PM
22upon7
Now I know the rules about this one, But I can't work out the second half of the sum,

1st half:
1. Find the value of $c$ such that the line with equation $y=2x+c$ is a tangent to the parabola with equation $y=x^2+3x$

$2x+c = x^2+3x$
$x^2+x-c=0$

the determinant = 0 then there is only one intersection (tangent)

through DOPS

$x^2+x+\frac{1}{4}=0$
$(x-\frac{1}{2})^2=0$

therefore $c = \frac{-1}{4}$

Now this is the part, I don't understand,

2. Find the possible values of c such that the line with equation
$y=2x+c$ twice intersects the parabola with equation $y=x^2+3x$

Any help would be much appreciated,

Thanks,

22upon7
• Mar 22nd 2009, 12:02 AM
mr fantastic
Quote:

Originally Posted by 22upon7
Now I know the rules about this one, But I can't work out the second half of the sum,

1st half:
1. Find the value of $c$ such that the line with equation $y=2x+c$ is a tangent to the parabola with equation $y=x^2+3x$

$2x+c = x^2+3x$
$x^2+x-c=0$

the determinant = 0 then there is only one intersection (tangent)

through DOPS

$x^2+x+\frac{1}{4}=0$
$(x-\frac{1}{2})^2=0$

therefore $c = \frac{-1}{4}$

Now this is the part, I don't understand,

2. Find the possible values of c such that the line with equation $y=2x+c$ twice intersects the parabola with equation $y=x^2+3x$

Any help would be much appreciated,

Thanks,

22upon7

You require the discriminant to be greater than zero.
• Mar 22nd 2009, 10:31 PM
22upon7
Thanks, so would that be proved like this:

$1-4c=0$

$1=4c$

$\frac{1}{4}=c$

Therefore $c>-\frac{1}{4}$
• Mar 23rd 2009, 12:32 AM
mr fantastic
Quote:

Originally Posted by 22upon7
Thanks, so would that be proved like this:

$1-4c=0$

$1=4c$

$\frac{1}{4}=c$

Therefore $c>-\frac{1}{4}$

The discriminant of $x^2 + x - c$ is not $1 - 4c$.

Once you calculate correct discriminant you need to solve the following inequality for c:

discriminant > 0.