Question

If zeroes of the polynomial 6xsquare + 4x+2a are alpha & 1/alpha then find the value of a

Answer 1

Concept:

The sum of zeros of a quadratic expression is equal to the negative value of the ratio of coefficient of variable with power 1 and coefficient of variable with power 2.

The Product of zeros of a quadratic expression is equal to the value of the ratio of constant term and coefficient of variable with power 2.

For example if the quadratic expression is $ax^2+bx+c$ and the zeros of this equation are $m,n$ then,

m + n = -b/a

mn = c/a

Given:

Given that the zeroes of the polynomial $6x^2+ 4x+2a$ are $\alpha,\frac{1}{\alpha}$.

Find:

The value of $a$.

Solution:

Given the quadratic expression is $6x^2+ 4x+2a$

And the zeros are $\alpha,\frac{1}{\alpha}$

So now, the product is,

$\alpha.\frac{1}{\alpha}=\frac{2a}{6}$

$\frac{a}{3}=1$, eliminating the like wise terms

$a=3$

Hence the value of a is given by 3.

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Answer 2

Answer:

3 is the required value of a

Step-by-step explanation:

Explanation:

Given that, $6x^2 + 4x + 2a$

$\alpha \ and \ \frac{1}{\alpha }$ are the zeroes of the polynomial.

As we know,

sum of zeroes of the polynomial ($\alpha + \beta$ )= $\frac{-b}{a}$

And the product of the zeroes ($\alpha \beta$) = $\frac{c}{a}$

Where a, b, and c are the coefficient of $x^2, x and \ constant \ value$

Step 1:

We have, $6x^2 + 4x + 2a$

a = 6 , b = 4 and c = 2a

Now, a product of zeroes ($\alpha .\frac{1}{\alpha }$) = $\frac{c}{a}$

⇒ 1 = $\frac{2a}{6}$

⇒ 6 = 2a

⇒ a = 3

Final answer:

Hence, 3 is the required value of a.

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