Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS

Question

My thinking:

Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$

By the AGM (Arithmetic-Geometric Mean Inequality):

We have

$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$

$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$

$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$

$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$

$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$

Substituting in the values for $x_1$ and $x_2$ we get:

$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$

$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$

$ = 0\le a^2-4a+4$

It seems as if I walked in circles through this process, can anyone help? Thanks in advance!

Answer 1

Another way to go... From $x^2 - a x + (a-1)$, we know that any root satisfies $$ x^2 = ax - (a-1) \text{,} $$ so \begin{align*} x_1^2 + x_2^2 &= (ax_1 - (a-1)) + (ax_2 - (a-1)) \\ &= a (x_1 + x_2) - 2a + 2 \text{.} \end{align*}

From \begin{align*} x^2 -ax +(a-1) &= (x-x_1)(x-x_2) \\ &= x^2 -(x_1 + x_2)x + x_1x_2 \text{,} \\ \end{align*} we have $x_1 + x_2 = a$, so \begin{align*} x_1^2 + x_2^2 &= a(a) - 2a + 2 \\ &= a^2 - 2a + 1 - 1 + 2 \\ &= (a-1)^2 + 1 \text{,} \end{align*} which is a nonnegative term plus $1$, so is minimized when the nonnegative term is zero, that is, when $a = 1$.

Answer 2

you have that: $$(x-x_1)(x-x_2)=0$$ since these are the roots, so: $$x^2-(x_1+x_2)x+x_1x_2=0$$ so: $$x_1+x_2=a\qquad x_1x_2=a-1$$ now we can say: $$(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=(x_1^2+x_2^2)+2(a-1)=a^2$$ which finally gives us: $$y=x_1^2+x_2^2=a^2-2a+2$$ now to find the minimum differentiate: $$\frac{dy}{da}=2a-2$$ so for $y'=0$ we get $a=1$ giving $\min(x_1^2+x_2^2)=1$

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In case you are not well versed on the use of derivatives, here is an alternative: $$\begin{align}y=&a^2-2a+2\\ =&(a-1)^2+1\end{align}$$ from this we can see that the minimum of the function occurs where $(a-1)=0\Rightarrow a=1$

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Also, here is a graph to visualise what is going on

Answer 3

$x_1^2+x_2^2=(x_1+x_2)^2-2(x_1x_2)=a^2-2(a-1)$
The last equility is Vieta or if you wish $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+(x_1x_2).$
So the min is when $a=1$ (because of usual calculus).

Answer 4

It looks like we are asked to notice that $ \ x^2 - ax + (a-1) \ = \ (x - 1)·( \ x - [a - 1] \ ) \ = \ 0 \ \ , $ so that one solution to the quadratic equation is always $ \ x_1 \ = \ 1 \ \ , $ with the other being $ \ x_2 \ = \ a - 1 \ \ . $ The function in question is then $ \ x_1^2 + x_2^2 \ = \ 1 + (a - 1)^2 \ \ , $ which will attain its minimum for $ \ a - 1 \ = \ 0 \ \Rightarrow \ a = 1 \ \ . $

As for sorting this out by the use of inequalities, that might work better with the RMS-GM inequality:

$$ \sqrt{\frac{x_1^2 \ + \ x_2^2}{2}} \ \ \ge \ \ \sqrt{x_1·x_2} \ \ \ge \ \ 0 \ \ \ \ \Rightarrow \ \ x_1^2 \ + \ x_2^2 \ \ \ge \ \ 2·x_1·x_2 \ \ = \ \ 2·(a \ - 1) \ \ \ge \ \ 0 \ \ , $$ where we have used the Viete relation for the product of the zeroes.