Let the root be 1 be $x-1=0$
Let it be called condition 1: $f(2)$ +$ f(4) $= $f(1)$ , not f(x).
For a Q like this , I can assume $ax^2$ + bx + c=0 where $a , b , c$ can be negative , +ve or even a bigger value for example ($9$ or$ 9+a$).
Q1 :Condition 1 can be true for a lot of polynomials and not just one. How do we decide upon which quadratic expression to choose and make a formula which is applicable for all the values.
One suggestion I took is $ax+b = 0 $since that is true. But this expression is true for all x. $ax+b $means that 1 and the other root , both are considered.
So , I considered.
$a(4) $+ $b(2) $+ $c $+ $a(16) $+ $b(4) $+ c = $1(a)$ + $1(b) $+ $c $.
Then , we get$ 5b $= $c-2c $+$ (1a – 20a)$.
b= $\frac{-c – 19a}{5}$ is till where I have solved
Putting b in -b/2a. Then , we get$\frac{5a+c}{10a}$
But in textbook , it is ax – $\frac{7a}{2}$
Best Answer
Let $f(x)=ax^2+bx+c$, if $x=1$ is a root then $a+b+c=0~~~~~(1)$ $f(2)+f(4)=0 \implies 10a+3b+c=0~~~~(2)$. (1) and (2) give $b=-9a/2, c=7a/2$ so the quadratic is $ax^2-9ax/2-9a/2=0 \implies 2x^2-9x+7=0 \implies x=\frac{9\pm\sqrt{81-56}}{4}=7/2,1.$ So the other root is $7/2$.