One of the zero is 2+root5 and sum of zeros is 4. Form a quadratic equation from the given information
Answers
Answered by
9
Hi ,
Let m , n are two roots of the quadratic
equation ,
m = 2 + √5 ,
m + n = 4
2 + √5 + n = 4
n = 4 - 2 - √5
n = 2 - √5 ,
Now ,
sum of the roots = m + n = 4
product of the roots = mn
= ( 2 + √5 ) ( 2 - √5 )
= 2² - ( √5 )²
= 4 - 5
= -1
Required quadratic equation ,
x² - ( m + n )x + mn = 0
x² - 4x - 1 = 0
I hope this helps you.
: )
Let m , n are two roots of the quadratic
equation ,
m = 2 + √5 ,
m + n = 4
2 + √5 + n = 4
n = 4 - 2 - √5
n = 2 - √5 ,
Now ,
sum of the roots = m + n = 4
product of the roots = mn
= ( 2 + √5 ) ( 2 - √5 )
= 2² - ( √5 )²
= 4 - 5
= -1
Required quadratic equation ,
x² - ( m + n )x + mn = 0
x² - 4x - 1 = 0
I hope this helps you.
: )
abhi569:
Thank you so much sir
Answered by
6
Hiii friend,
Let Alpha = 2+✓5
Sum of zeros = (Alpha × Beta)
4 = 2+✓5 + Beta
Beta = 4-2-✓5 => 2-✓5
Product of Zeros = (Alpha × Beta) = (2+✓5)(2-✓5) = (2)² - (✓5)² = 4-5 = -1.
Therefore,
Required POLYNOMIAL = X²-(Alpha+Beta)X+Alpha × Beta
=> X²-(4)X+(-1)
=> X²-4X-1
HOPE IT WILL HELP YOU....... :-)
Let Alpha = 2+✓5
Sum of zeros = (Alpha × Beta)
4 = 2+✓5 + Beta
Beta = 4-2-✓5 => 2-✓5
Product of Zeros = (Alpha × Beta) = (2+✓5)(2-✓5) = (2)² - (✓5)² = 4-5 = -1.
Therefore,
Required POLYNOMIAL = X²-(Alpha+Beta)X+Alpha × Beta
=> X²-(4)X+(-1)
=> X²-4X-1
HOPE IT WILL HELP YOU....... :-)
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