Question

find the zeros of quadratic polynomial X square + R X + 10 and verify the relationship between the zeros and the coefficient​

Answer 1

Answer:

Chapter 2: Rings and Polynomials ... the product of any two non-zero elements of R is itself non-zero. ... Lemma 2.5 Let R be a unital commutative ring, and let X be a subset of ... DRR11-12-Ia-b-5.

Answer 2

The zeroes of quadratic polynomial $x^2+RX+10$ are

-2 and -5

Step-by-step explanation:

Given equation:

$x^2+Rx+10$

Compare it with

$x^2-(\alpha+\beta)x+\alpha\beta$

We get

$\alpha+\beta=-R$

$\alpha\beta=10$

$10=2\times 5$

When the sum of zeroes is negative it means both zeroes are negative because the product of zeroes is positive.

$\alpha+\beta=-5-2=-7$

Therefore, we can take

$\alpha=-2,\beta=-5$

Substitute the values

$x^2-(-7)x+10=x^2+7x+10$

Compare it with

$ax^2+bx+c$

We get a=1,b=7,c=10

Sum of zeroes=-7

$-\frac{b}{a}=-\frac{7}{1}=-7$

Sum of zeroes=$-\frac{b}{a}$

Product of zeroes =10

$\frac{c}{a}=\frac{10}{1}=10$

Product of zeroes =$\frac{c}{a}$

Hence, the relation between the zeroes and the coefficient is verified.

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https://brainly.in/question/9844754:Answered By Antareepray