Question

P(x) =x^2-5
Find zeroes of polynomials and verify relation between zeroes and coefficient

Answer 1

Step-by-step explanation:

Given -

• p(x) = x² - 5

To Find -

• The zeroes of the polynomial and verify the relationship between the zeroes and the coefficient

Now,

→ x² - 5

→ (x)² - (√5)²

Using the identity of a² - b² = (a + b)(a - b), we get :

→ (x + √5)(x - √5)

Zeroes are -

→ x + √5 = 0 and x - √5 = 0

→ x = -√5 and x = √5

Verification :-

As we know that :-

• α + β = -b/a

→ -√5 + √5 = -(0)/1

→ 0 = 0

LHS = RHS

And

• αβ = c/a

→ -√5 × √5 = -5/1

→ -5 = -5

LHS = RHS

Hence,

Verified...

It shows that our answer is absolutely correct.

Answer 2

✯✯ QUESTION ✯✯

$P(x) ={x}^{2}-5$

Find zeroes of polynomials and verify relation between zeroes and coefficient..

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✰✰ ANSWER ✰✰

☛Here ,

$\longrightarrow{{x}^{2}-5}$

$\longrightarrow{{(x)}^{2}-\sqrt{(5)}^{2}}$

☛Using Identity : -

$\longrightarrow{{a}^{2}-{b}^{2}=(a+b)(a-b)}$

$\longrightarrow{(x+\sqrt{5)} (x-\sqrt{5)}}$

• $x=-\sqrt{5}$
• $x=\sqrt{5}$

➥So, -√5 and √5 are zeroes of The Polynomial x²-5.

☛Sum of Zeroes : -

$\longrightarrow{\alpha\beta=\dfrac{-b}{a}}$

$\longrightarrow{-\sqrt{5}+\sqrt{5}=\dfrac{-(0)}{1}}$

$\longrightarrow{0=0}$

$\longrightarrow{\small{\boxed{\bold{\bold{\pink{\sf{L.H.S=R.H.S}}}}}}}$

☛Product of Zeroes : -

$\longrightarrow{\alpha\beta=\dfrac{c}{a}}$

$\longrightarrow{-\sqrt{5}\times\sqrt{5}=\dfrac{-5}{1}}$

$\longrightarrow{-5=-5}$

$\longrightarrow{\small{\boxed{\bold{\bold{\orange{\sf{L.H.S=R.H.S}}}}}}}$