If the length of three sides of $$ riangle ABC$$ 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 $$ riangle ABC$$ is a(n)

Question

If the length of three sides of $$	riangle ABC$$ 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 $$	riangle ABC$$ is a(n)

Uniteasy Admin · Accepted Answer

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).

For given quadratic function, 
$$y=(a+b)x^2+2cx-(a-b)$$n
the function for axis of symmetry is
$$x =  \dfrac{-2c}{2(a+b)}  = -  \dfrac{1}{2}  $$
Then,
$$a+b=2c$$n
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,
$$c=3b-2a$$n
Substituting (3) to (2) results in
$$ a+b= 6b-4a$$
$$a=b$$n
Substituting (4) to (2) yields
$$ a=c$$n
Therefore
$$a=b=c$$
$$	riangle ABC$$ is an equilateral triangle

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