find the polynomial whose zeroes are
a) 3+ underroot 2 and 3 - underroot 2
Also show the working.
Answers
Answered by
5
Hey there!
Sum of roots = 3 + √2 + 3 - √2 = 6 .
Product of roots = ( 3 + √2 ) (3 - √2 ) = 9 - 2 = 7 .
Now,
We know that,
Quadratic equation =>
x² - ( sum of roots) x + product of roots = 0 .
x² - ( 6)x + 7 = 0 .
Hence, x²-6x+7 = 0 is the required quadratic equation.
Sum of roots = 3 + √2 + 3 - √2 = 6 .
Product of roots = ( 3 + √2 ) (3 - √2 ) = 9 - 2 = 7 .
Now,
We know that,
Quadratic equation =>
x² - ( sum of roots) x + product of roots = 0 .
x² - ( 6)x + 7 = 0 .
Hence, x²-6x+7 = 0 is the required quadratic equation.
Answered by
1
Answer:
Sum of roots = 3 + √2 + 3 - √2 = 6 .
Product of roots = ( 3 + √2 ) (3 - √2 ) = 9 - 2 = 7 .
Now,
We know that,
Quadratic equation =>
x² - ( sum of roots) x + product of roots = 0 .
x² - ( 6)x + 7 = 0 .
Hence, x²-6x+7 = 0 is the required quadratic equation.
Step-by-step explanation:
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