Let f(x) be a quadratic polynomial satisfying

Henry Winters

Henry Winters

Answered question

2022-04-13

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.

Answer & Explanation

oanhtih6

oanhtih6

Beginner2022-04-14Added 10 answers

Let f(x)=ax2+bx+c, if x=1 is a root then a+b+c=0 (1)
f(2)+f(4)=0⇒10a+3b+c=0≈≈(2)
(1) and (2) give b=−9a2,c=7a2 so the quadratic is ax2−9ax2−9a2=0⇒2x2−9x+7=0⇒x=9±81−564=72,1.
So the other root is 72.
Videoad3u

Videoad3u

Beginner2022-04-15Added 15 answers

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=1+r2, 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 ⇒ k = −a·(1−h)2  .
 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
⇒  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]
⇒  (4−4h+h2) + (16−8h+h2) − 2·(1−2h+h2)  =  18 − 8h  =  0
⇒  h  =  94  ⇒  r  =  2h − 1  =  92 − 1  =  72  .
The family of quadratic polynomials satisfying the specified conditions is then  a·(2x2−9x+7)  .

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