Question

if -2 and 3 are the zeroes of the polynomial ax2+bx-6 then find the value of a and b​

Answer 1

Step-by-step explanation:

Given:-

polynomial. f(x)= ax^2+bx-6

zeroes are -2 and 3

value of a and b=?

solution:-

f(-2)=4a-2b-6=0

4a-2b=6. (1)

f(3)=9a+3b-6=0

9a+3b=6. (2)

solving (1) and (2)

3*(1)+2*(2)

12a-6b=18

+18a+6b=12

a= 1

b= -1

Answer 2

Concept

a polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Given

The given polynomial is $ax^2+bx-6$ whose zeroes are $-2$ and $3.$

Find

We have to seek out the worth of $a$ and $b.$

Solution

The given polynomial is $ax^2+bx-6$.

The general variety of an algebraic polynomial is $ax^2+bx+c=0$.

Firstly, we are going to compare the final polynomial with given polynomial, we get

$a=a,b=b,c=-6$

As we all know that the sum of the zeroes of the polynomial are $\alpha+\beta=-\frac{b}{a}$ and therefore the product of zeroes are $\alpha \beta=\frac{c}{a}$.

Here, $\alpha=-2$ and $\beta=3$.

Now, we'll substitute these values within the sum of zero, we get

$\begin{aligned}-2+3&=-\frac{b}{a}\\ 1&=-\frac{b}{a}\\ a&=-b\end$                                 ......(1)

Further, we'll substitute these values in product of zero, we get

$\begin{aligned}(-2)\times (3)&=\frac{-6}{a}\\ -6&=\frac{-6}{a}\end$

Furthermore, we'll cross multiply each side and simplify, we get

$\begin{aligned}a\times (-6)&=-6\\ a&=1\end$

Substitute this value in equation (1), we get

$\begin{aligned}1&=-b\\ b&=-1\end$

Hence, the worth of $a$ and $b$ is $1$ and $-1.$

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