Reason for repeated root while solving tangent and curve.

Question

I was solving a question from a book which is stated as: Let a tangent be drawn to $y=x^4-16x^2$ at $x = 1$. I need to find all point of intersection between curve and the tangent.

I started by finding equation of tangent. Using slope point form , equation of tangent: $y=-28x+13$.

Now equating $y$ of the line in the curve, I get a biquadratic equation in $x$, which is: $x^4-16x^2+28x-13=0$

Now the solution in the book states that $x=1$ is a repeated root of the above equation, since it satisfies both curve and the tangent. I don't understand this, by this logic won't we get a repeated root in every time when we solve equation of tangent and curve since the point of tangency satisfies both curve and tangent?

Can anybody provide a better reason why do I get a repeated root?

Answer 1

Taking the biquadratic you got $$P(x)=x^4-16x^2+28x-23=0$$

It is easy to check that $$P(1)=1-16+28-13=0$$

thus $$P(x)=(x-1)Q(x)$$ with $ deg(Q)=3$.

But $$P'(x)=Q(x)+(x-1)Q'(x)$$ $$=4x^3-32x+28$$

Now, observe that

$$P'(1)=4-32+28=0=Q(1)$$

thus $$Q(x)=(x-1)R(x)$$ with $deg(R)=2$. and finally $$P(x)=(x-1)^2R(x)$$

$x=1 $ is repeated root of $ P(x)=0.$

Answer 2

But we do get a double root every time we solve such a problem. Suppose that you were working with $x=2$ instead of $x=1$. Then the tangent line is $y=16-32x$ and $2$ is a double root of $x^4-16x^2+32x-16$, since$$x^4-16x^2+32x-16=(x-2)^2(x^2+4x-4).$$

Answer 3

Let the line $y = g(x)$ be tangent to the curve $y = f(x)$ at $P = (a, f(a))$. Define $h(x) = f(x) - g(x)$. By definition $h(a) = 0$ since $f(x)$ and $g(x)$ intersect at $P$. Also since $g(x)$ is tangent to $f(x)$ at $P$, $g'(a) = f'(a) \implies h'(a) = 0$. Since $h(a) = h'(a) = 0$, $a$ is a double root of $h(a)$