Let f(x) be a quadratic polynomial satisfying f(2) + f(4) = 0. If unity is one root of f(x) = 0 then find the other root.

Question

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 - $\frac{7a}{2}$

Answer 1

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 $7/2$.

Answer 2

Tavish's method of working from the factored form of the polynomial is certainly the most direct. We can also approach this by using the "vertex form" of the polynomial, $ \ f(x) \ = \ a·(x - h)^2 + k \ \ . $ We know that the axis of symmetry of the corresponding parabola is halfway between its $ \ x-$intercepts (the zeroes of the polynomial), so we have $ \ h \ = \ \frac{1 + r}{2} \ \ , \ \ r \ $ being the unknown zero (we will return to this shortly).

The information about $ \ x \ = \ 1 \ $ tells us that $ \ f(1) \ = \ a·(1 - h)^2 + k \ = \ 0 \ \Rightarrow \ k \ = \ -a·(1 - h)^2 \ \ . $ Since $ \ f(2) \ = \ -f(4) \ \ , $ the Intermediate Value Theorem leads us to expect that $ \ 2 < r < 4 \ \ . $ We may write $$ a·(2 - h)^2 \ - \ a·(1 - h)^2 \ \ = \ \ -a·(4 - h)^2 \ + \ a·(1 - h)^2 $$ $$ \Rightarrow \ \ a·(2 - h)^2 \ + \ a·(4 - h)^2 \ - \ 2·a·(1 - h)^2 \ \ = \ \ 0 $$ [it is clear by now (if it wasn't already) that $ \ a \ $ may have any non-zero value] $$ \Rightarrow \ \ (4 - 4h + h^2) \ + \ (16 - 8h + h^2) \ - \ 2· (1 - 2h + h^2) \ \ = \ \ 18 \ - \ 8h \ \ = \ \ 0 $$ $$ \Rightarrow \ \ h \ \ = \ \ \frac94 \ \ \Rightarrow \ \ r \ \ = \ \ 2h \ - \ 1 \ \ = \ \ \frac92 \ - \ 1 \ \ = \ \ \frac72 \ \ . $$ The family of quadratic polynomials satisfying the specified conditions is then $ \ a·(2x^2 - 9x + 7) \ \ . $