MHB Proving $1 \leq a \leq 9$ for Quadratic Equations

[Albert1]

$a^2-bc-8a+7=0$
$b^2+c^2+bc-6a+6=0$
prove:$1\leq a\leq9$

 

[anemone]

My solution:

Spoiler

I cheated a bit because my solution is sort of working backwards, since we know we have to prove that $1\le a \le 9$.

Since $(a-1)(a-9)=a^2-10a+9$, we manipulate the first equation algebraically to get:

$a^2-10a+9+2a-2-bc=0$

$a^2-10a+9=bc-2(a-1)$(*)

Rewrite the second equation such that it becomes

$b^2+c^2+bc-6a+6=0$

$6(a-1)=(b-c)^2+3bc$

$2(a-1)=\dfrac{(b-c)^2}{3}+bc$

Substitute the above into (*) yields

$a^2-10a+9=bc-\left(\dfrac{(b-c)^2}{3}+bc \right)=-\dfrac{(b-c)^2}{3}\le 0$

$\therefore a^2-10a+9 \le 0$ this shows $1\leq a\leq9$ and we're done.

 

[Albert1]

My solution:

Spoiler

I cheated a bit because my solution is sort of working backwards, since we know we have to prove that $1\le a \le 9$. Since $(a-1)(a-9)=a^2-10a+9$, we manipulate the first equation algebraically to get: $a^2-10a+9+2a-2-bc=0$ $a^2-10a+9=bc-2(a-1)$(*) Rewrite the second equation such that it becomes $b^2+c^2+bc-6a+6=0$ $6(a-1)=(b-c)^2+3bc$ $2(a-1)=\dfrac{(b-c)^2}{3}+bc$ Substitute the above into (*) yields $a^2-10a+9=bc-\left(\dfrac{(b-c)^2}{3}+bc \right)=-\dfrac{(b-c)^2}{3}\le 0$ $\therefore a^2-10a+9 \le 0$ this shows $1\leq a\leq9$ and we're done.

very good ! working backwards is also a nice method,it gives us a hint to work forwards.

 

[kaliprasad]

$a^2-bc-8a+7=0$
$b^2+c^2+bc-6a+6=0$
prove:$1\leq a\leq9$

Spoiler

we have from 2nd equation
$b^2+c^2+bc = 6a- 6\ \cdots (1)$
from 1st equation
$bc = a^2-8a + 7\ \cdots (2)$
multiply (2) by 3 and subtract from (1)
$(b-c)^2 = 6a-6 - 3(a^2- 8a +7)$
=$-3a^2+30a-27$
=$-3(a^2-10a+9)$
so $a^2-10a + 9 \le 0$
hence $1 \le a \le 9$

 

[Albert1]

Spoiler

we have from 2nd equation
$b^2+c^2+bc = 6a- 6\ \cdots (1)$
from 1st equation
$bc = a^2-8a + 7\ \cdots (2)$
multiply (2) by 3 and subtract from (1)
$(b-c)^2 = 6a-6 - 3(a^2- 8a +7)$
=$-3a^2+30a-27$
=$-3(a^2-10a+9)$
so $a^2-10a + 9 \le 0$
hence $1 \le a \le 9$

very nice !