Question

If p(x) = ax2 + bx + c is a quadratic polynomial then what is relation of c/a with zeroes of p(x)?

Answer 1

Answer:

If p(x) = ax² + bx + c

Then

Sum of the zeroes = - b/a

Product of the Zeros = c/a

So

c/a = Product of the Zeros of p(x)

Answer 2

Answer:

Relation between $\frac{c}{a}$ with zeroes of p(x) is ,

$\frac{c}{a}$ = product of zeroes of the polynomial p(x)

Step-by-step explanation:

Explanation:

Given , p(x) = $ax^2 +bx +c$

The negative of the coefficient of x by the coefficient of $x^2$ is equal to the total of the zeroes. The constant term divided by the coefficient of $x^2$ is equal to the product of the zeroes.

Let $\alpha \ and \ \beta$ be the zeroes of the given polynomial .

Step 1:

We have p

p(x) = $ax^2 +bx +c$

So , coefficient of $x^2$ is a , coefficient of x is b and constant term is c .

Sum of zeroes of polynomial p(x) ($\alpha +\beta$) =   $\frac{-coefficient \ of x }{coefficient \ of \ x^2}$=  $\frac{-b}{a}$

Product of  zeroes of polynomial p(x) ($\alpha \beta$) =$\frac{constant \ term }{ coefficient \ of x^2}$=  $\frac{c}{a}$

Final answer :

Hence , $\frac{c}{b}$ is equal to product of zeroes of polynomial p(x).

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