Find the zeroes of p(x)=2x^2-(1+2√2)x+2√2 by factorisation method.
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Zayer
1) 2s2 - (1 + 2√2) s + √2 = 2s2 - s - 2√2s + √2 = S(2s - 1) - √2(2s - 1) = (2s - 1)(s - √2) 2s - 1 = 0 ⇒ s = 1/2 s - √2 = 0 ⇒ s = √2 Therefore, Zeroes of the polynomial are 1/2 and √2.
If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then α +β = − b/a αβ = c/a. Therefore, sum of the roots is (1 + 2√2) / 2 i.e. (1/2 + √2) Product of the roots is √2 / 2 = 1/√2.
Answers & Comments
= S(2s - 1) - √2(2s - 1)
= (2s - 1)(s - √2)
2s - 1 = 0 ⇒ s = 1/2
s - √2 = 0 ⇒ s = √2
Therefore, Zeroes of the polynomial are 1/2 and √2.
If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then
α +β = − b/a
αβ = c/a.
Therefore, sum of the roots is (1 + 2√2) / 2 i.e. (1/2 + √2)
Product of the roots is √2 / 2 = 1/√2.