﻿ If the length of three sides of \triangle ABC is a,b,c, such that when x=-\dfrac{1}{2},

Multiple Choice Question (MCQ)

If the length of three sides of \triangle ABC is a,b,c, such that when x=-\dfrac{1}{2}, the min value of quadratic function y=(a+b)x^2+2cx-(a-b) is -\dfrac{b}{2}, then \triangle ABC is a(n)

1. equilateral triangle

2. ×

isosceles triangle

3. ×

right triangle

4. ×

acute triangle

Collected in the board: Quadratic function

Steven Zheng posted 5 months ago

1. For a standard quadratic function,

y=ax² +bx+c

the coordinates of the vertex of the curve is

\Big( \dfrac{-b}{2a},\dfrac{4ac-b^2}{4a} \Big) (a, b,c is not related to the given question here).

y=(a+b)x^2+2cx-(a-b)
(1)

the function for axis of symmetry is

x = \dfrac{-2c}{2(a+b)} = - \dfrac{1}{2}

Then,

a+b=2c
(2)

Substitute (2) to (1)

y=2cx^2+2cx-(a-b)

=2c(x+\dfrac{1}{2} )^2-\dfrac{c}{2} -(a-b)

Therefore

y_{min} = -\dfrac{c}{2} -(a-b) =-\dfrac{b}{2}

Then,

c=3b-2a
(3)

Substituting (3) to (2) results in

a+b= 6b-4a

a=b
(4)

Substituting (4) to (2) yields

a=c
(5)

Therefore

a=b=c

\triangle ABC is an equilateral triangle

Steven Zheng posted 5 months ago

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