Minimum value based on solutions

Algebra Level pending

Given an equation $2x^2-(6m-3)x-3m+1 = 0$ where $x$ and $m$ are real numbers and $m$ is the parameter.

The equation above has 2 distinct real solutions which are $x_1$ and $x_2$ .

Find $m$ so that the value of $A=x_1^2 + x_2^2$ is minimum.

The answer is in the form of $\frac{a}{b}$ where $a,b$ are positive coprime integers. Type $a+b$ .

1 solution

A Former Brilliant Member
Jul 25, 2020

$x_1+x_2=\dfrac {6m-3}{2},x_1x_2=-\dfrac {3m-1}{2}$

$\implies A=x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$

$=9\left (m^2-\dfrac 23 m+\dfrac {5}{36}\right )$

$=9(m-\frac 13)^2+\dfrac 14\geq \dfrac 14$

So, the minimum of $A$ is $\dfrac 14$ when $m=\dfrac 13$

Hence $a=1,b=3,a+b=\boxed 4$ .