There are two quadratic functions $f(x),~g(x)$ and a linear function $h(x).$ $f(x)$ and $h(x)$ contact with each other at only one point, $(\alpha,~f(\alpha)).$ $g(x)$ and $h(x)$ contact with each other at only one point, $(\beta,~h(\beta)).$

They satisfy the below conditions:

- Leading coefficients of $f(x)$ and $g(x)$ are $1$ and $4,$ respectively.
- Two positives $\alpha$ and $\beta$ satisfy $\alpha:\beta=1:2.$

Let $t$ be the $x$ -coordinate of the point of intersection of $y=f(x)$ and $y=g(x)$ whose $x$ -coordinate is in between $\alpha$ and $\beta.$

Find the value of $\dfrac{210t}{\alpha}.$

*
This problem is a part of
<Grade 10 CSAT Mock test> series
.
*

The answer is 350.

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$f(x)-h(x)=0$ has an equal root of $\alpha,$ and since the leading coefficient of $f(x)$ is $1,$

$f(x)=(x-\alpha)^2+h(x).$

Using similar method, we get $g(x)=4(x-\beta)^2+h(x).$

Know that $\beta=2\alpha.$

Then let's solve $f(t)=g(t).$

$(t-\alpha)^2+h(t)=4(t-\beta)^2+h(t)$

$3t^2-14\alpha t+15\alpha^2=0$

$(3t-5\alpha)(t-3\alpha)=0.$

Since $\alpha<t<2\alpha,$ we get $t=\dfrac{5}{3}\alpha.$

$\therefore~\dfrac{210t}{\alpha}=\boxed{350}.$