Question

write the sum of zeros of a quadratic polynomial p(x)= ax2 + bx + c where ,a is not equal to zero​

Answer 1

Answer:

Hey dear !!!

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==> In the equation ,

p(x) = ax² + bx + c

We have to find the sum of zeroes and product of zeroes .

We have,

a = a

b = b

c = c

Let, α and β be the zeros of the given polynomial,

We know that,

Sum of zeroes = α + β = -b/a

= -b/a

Also we know,

Product of zeroes = αβ = c/a

= c/a

Therefore, -b/a and c/a are the sum and product of the given polynomial .

Answer 2

$-\dfrac{b}{a}$

The sum of the zeroes of a quadratic polynomial $p(x)=ax^{2}+bx+c$ where $a$ is not equal to zero, is $(-\dfrac{b}{a})$.

Step-by-step explanation:

The given polynomial is

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

Let $\alpha$ and $\beta$ be the zeroes of $p(x)$. Then

$a\alpha^{2}+b\alpha+c=0$ ... ... (1)

$a\beta^{2}+b\beta+c=0$ ... ... (2)

Now, (1) - (2) gives

$a(\alpha^{2}-\beta^{2})+b(\alpha-\beta)=0$

$\Rightarrow a(\alpha+\beta)(\alpha-\beta)+b(\alpha-\beta)=0$

$\Rightarrow (\alpha-\beta)\{a(\alpha+\beta)+b\}=0$

Either $\alpha-\beta=0$ or, $a(\alpha+\beta)+b=0$

So, $a(\alpha+\beta)+b=0$

$\Rightarrow a(\alpha+\beta)=-b$

$\Rightarrow \alpha+\beta=-\dfrac{b}{a}$, where $a\neq 0$

This shows that the sum of the zeroes of the given polynomial is $(-\dfrac{b}{a})$.

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