Math, asked by shivraj7562, 4 months ago

Find the zeroes of the quadratic polynomial x2-2√2x and verify the relationship between the zeroes and the coefficients.​

Answers

Answered by Arceus02
11

Given:-

  • p(x) = x² - (2√2)x

To find:-

  • Zeroes of p(x) and verify the relationship between the zeroes.

Answer:-

A zero/root of a polynomial p(x) is that value at which the value of the polynomial becomes 0.

So, if x₁ and x₂ are the zeroes of p(x), then p(x₁) = p(x₂) = 0

Given polynomial,

x² - 2√2x = 0

Taking x as common,

→ x(x - 2√2) = 0

Either x = 0, or (x - 2√2) = 0

1st situation:

x₁ = 0

(first zero)

2nd situation:

x - 2√2 = 0

→ x₂ = 2√2

(second zero)

Verify:-

Considering p(x) = x² - (2√2)x with the standard form, that is, ax² + bx + c, we get,

  • a = 1
  • b = -2√2
  • c = 0

We know that, for a polynomial ax² + bx + c having roots x₁ and x₂,

  • x₁ + x₂ = -b/a
  • x₁x₂ = c/a

Here,

x₁ + x₂ = 0 + 2√2

→ 2√2 = -b/a

→ 2√2 = -(-2√2)/1

→ 2√2 = 2√2

Hence verified!

x₁x₂ = 0 * 2√2

→ 0 = c/a

→ 0 = 0/1

→ 0 = 0

Hence verified!

Answered by FantasyWorld2
1

Given:-

p(x) = x² - (2√2)x

To find:-

Zeroes of p(x) and verify the relationship between the zeroes.

Answer:-

A zero/root of a polynomial p(x) is that value at which the value of the polynomial becomes 0.

So, if x₁ and x₂ are the zeroes of p(x), then p(x₁) = p(x₂) = 0

Given polynomial,

x² - 2√2x = 0

Taking x as common,

→ x(x - 2√2) = 0

Either x = 0, or (x - 2√2) = 0

1st situation:

x₁ = 0

(first zero)

2nd situation:

x - 2√2 = 0

→ x₂ = 2√2

(second zero)

Verify:-

Considering p(x) = x² - (2√2)x with the standard form, that is, ax² + bx + c, we get,

a = 1

b = -2√2

c = 0

We know that, for a polynomial ax² + bx + c having roots x₁ and x₂,

α+β = -b/a

αβ = c/a

Here,

α+β= 0 + 2√2

→ 2√2 = -b/a

→ 2√2 = -(-2√2)/1

→ 2√2 = 2√2

Hence verified!

αβ = 0 * 2√2

→ 0 = c/a

→ 0 = 0/1

→ 0 = 0

Hence verified!

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