Find the zeros of the polynomial x²-2√a x+(a-b)???
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Answers
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 !!
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Answer:
√a+√b,√a-√b
Step-by-step explanation:
We know that the zero of a quadratic polynomial=[-b±√(b²-4ac)]/2a
From this, we know that Alpha=[-b+√(b²-4ac)]/2a & Beta=[-b-√(b²-4ac)]/2a
From substituting the respective values, we get
Alpha=[-(-2√a)+√(4a-4a-4b)]/2
=>2(√a+√b)/2
=>√a+√b
Therefore, we get Beta=√a-√b
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