1. Simultaneous Linear and Quadratic Equations

Can someone please show me how to work through this type of sum?

Find the value(s) of $a$ such that the line with equation $y=x$ is tangent to the parabola with equation $y=x^2+ax+1$

I know the determinant should equal zero to create a tangent with the parabola, but I don't know how to solve it.

Any help will be much appreciated,

Thanks,

22upon7

2. Originally Posted by 22upon7
Can someone please show me how to work through this type of sum?

Find the value(s) of $a$ such that the line with equation $y=x$ is tangent to the parabola with equation $y=x^2+ax+1$

I know the determinant should equal zero to create a tangent with the parabola, but I don't know how to solve it.

Any help will be much appreciated,

Thanks,

22upon7
$x = x^2 + ax + 1 \Rightarrow x^2 + (a - 1)x + 1 = 0$.

You should know that the discriminant of $Ax^2 + Bx + C = 0$ is $B^2 - 4AC$.

Therefore ....

3. Originally Posted by 22upon7
Can someone please show me how to work through this type of sum?

Find the value(s) of $a$ such that the line with equation $y=x$ is tangent to the parabola with equation $y=x^2+ax+1$

I know the determinant should equal zero to create a tangent with the parabola, but I don't know how to solve it.
I do not know what 'sum' you are referring to. I also do not know what determinant you are referring to either. Are you referring to the discriminant in the quadratic formula?

We require the two functions to intersect at one and only one point. Setting the y values equal to each other we get,
$x=x^2 +ax+1 \implies 0 = x^2+(a-1)x+1$
In order for this quadratic equation to yield one solution (intersection point) we require the discriminant (the part inside the radical in the quadratic equation) to be zero.
$(a-1)^2 - 4*1*1 = 0 \implies (a-1)^2 = 4 \implies a = 1 \pm 2 \implies a = 3, -1$
hth