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