Question

find out the sum and the multiplication of the root of the polynomial:
$a( {x }^{2} + 1) - x( {a}^{2} + 1)$
​

Answer 1

$\large{\underline{\bf{\green{Given:-}}}}$

✰ p(x) = $a( {x }^{2} + 1) - x( {a}^{2} + 1)$

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰we need to find the relationship between the zeroes and coefficients.

$\huge{\underline{\bf{\red{Solution:-}}}}$

$: \mapsto \:$a(x² + 1) - x(a² + 1)

$: \mapsto \sf\:$ax² - xa² + a - x

$: \mapsto \sf\:$ xa (x - a) -1(x - a)

$: \mapsto \sf\:$(xa - 1)(x - a)

So,

The zeroes of the given polynomial:-

$: \mapsto \sf\:$xa - 1 = 0

$: \mapsto \sf\:$ xa = 1

$: \mapsto \sf\:$ x = 1/a

Or

$: \mapsto \sf\:$x - a = 0

$: \mapsto \sf\:$x = a

Now,

Let α and β are the zeroes of the given polynomial.

Let α = 1/a

and β = a

Sum of zeroes = α + β

$: \mapsto \sf\:$ 1/a + a

$: \mapsto \sf\:\frac{1+a^2}{a}$

Product of zeroes:- αβ

$: \mapsto \sf\:$1/a × a

$: \mapsto \sf\:$ 1

Now,

Verification:-

p(x) = ax² -(a² + 1)x + a

sum of zeroes:-

α + β = - b/a

$: \mapsto \sf\:\frac{1+a^2}{a}=\frac{-(-a^2)+1}{a}$

$: \mapsto \sf\:\frac{1+a^2}{a}=\frac{1+a^2}{a}$

Product of zeroes:-

αβ = c/a

$: \mapsto \sf\:$1/a × a = a/a

$: \mapsto \sf\:$ 1 = 1

LHS = RHS

hence , relationship is verified.

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Answer 2

Step-by-step explanation:

Given -

• p(x) = a(x² + 1) - x(a² + 1)

To Find -

• Sum and multiplication of the root of the polynomial.

Now,

→ a(x² + 1) - x(a² + 1) = 0

→ ax² + a - xa² - x = 0

→ ax² - xa² - x + a = 0

→ ax(x - a) - 1(x - a)

→ (ax - 1)(x - a)

Zeroes are :-

→ ax - 1 = 0 and x - a = 0

→ x = 1/a and x = a

Then,

The sum of the root of the polynomial is

→ 1/a + a

→ a² + 1/a

And

The product of the root of the polynomial is

→ 1/a × a

→ 1

Hence,

The sum of the root of the polynomial is 1 + a²/a

And

The product of the root of the polynomial is 1