6. Find the quadratic polynomial, the sum of whose zeroes is v2 and their product is 1/3.
Hence find the zeroes of the polynomial.
Answers
Given :- Find the quadratic polynomial, the sum of whose zeroes is √2 and their product is 1/3. Hence find the zeroes of the polynomial. ?
Solution :-
we know that, if sum and product of zeros are given , then,
- Quadratic Equation = x² - (Sum of the zeros)x + Product of zeros = 0 .
given,
- sum of zeros = √2
- product of zeros = (1/3) .
putting both values we get,
→ x² - √2x + (1/3) = 0
→ (3x² -3√2x + 1)/3 = 0
→ 3x² - 3√2x + 1 = 0 = Required Quadratic polynomial . (Ans.)
Now, we know that, Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,
- x = [ -b±√(b²-4ac) ] / 2a
comparing 3x² - 3√2x + 1 = 0 with ax² +bx + c = 0 we get,
- a = 3
- b = (-3√2)
- c = 1
therefore,
→ x = [ -b±√(b²-4ac) ]/ 2a
→ x = [-(-3√2) ± √{(-3√2)² - 4 * 3 * 1} ] / 2*3
→ x = [3√2 ± √(18 - 12)] / 6
→ x = [3√2 ± √6] / 6
hence,
→ x ∈ { (3√2 + √6)/6 , (3√2 + √6)/6 } . (Ans.)
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