#21 of Sept 2016 Grade 10 CSAT(Korean SAT) mock test

Algebra Level 3

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

They satisfy the below conditions:

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

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

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


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


The answer is 350.

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1 solution

Boi (보이)
Jul 23, 2017

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

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

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

Know that β = 2 α . \beta=2\alpha.


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

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

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

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

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

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

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