find the zeros of x2 + 2√2x - 6 a quadratic polynomial and verify the relationship between the zeros and their coefficients
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Hi there !!
p(x) = x² + 2√2x - 6
We can find the zeros by factorization [ by splitting the middle terms ]
x² + 2√2x - 6
x² + 3√2x - √2x - 6
x [ x + 3√2 ] - √2 [ x+ 3√2]
[ x - √2 ] [ x+ 3√2]
the zeros are = √2 and -3√2
α = √2
β = -3√2
a = 1
b = 2√2
c = -6
Sum of zeros = √2 + -3√2 = -2√2 = -b/a [ -b/a = -2√2 ]
Product of zeros = √2 × -3√2 = - 6 = c/a [c/a = -6/1 = -6]
Hope this helps you !!
p(x) = x² + 2√2x - 6
We can find the zeros by factorization [ by splitting the middle terms ]
x² + 2√2x - 6
x² + 3√2x - √2x - 6
x [ x + 3√2 ] - √2 [ x+ 3√2]
[ x - √2 ] [ x+ 3√2]
the zeros are = √2 and -3√2
α = √2
β = -3√2
a = 1
b = 2√2
c = -6
Sum of zeros = √2 + -3√2 = -2√2 = -b/a [ -b/a = -2√2 ]
Product of zeros = √2 × -3√2 = - 6 = c/a [c/a = -6/1 = -6]
Hope this helps you !!
Anonymous:
here the middle term is 2root2
we have to chose two numbers such that , their sum = 2root2
and the product of the no:s = -6
we have learnt this from class 8th itself
ok
thnx a lot
welcome !
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