Asked by jhon mathew
If alpha and beta are the zeroes of the polynomial f (x) = 5y square -7y + 1. Find another polynomial whose zeroes are 2 alpha / beta and 2beta / alpha
Whose zeroes ar 1/ 2alpha + beta and 1/2beta + alpha
Whose zeroes ar 1/ 2alpha + beta and 1/2beta + alpha
Answers
Answered by
Jai
Let alpha = a
Let beta = b
Because it's easier to type a and b, than copy-pasting alpha and beta symbols.
Anyway, recall that for a quadratic equation in the form
y = ax^2 + bx + c
The sum and product of the roots can be determined as such:
y = x^2 - (sum)x + (product)
Of course, the numerical coefficient of x^2 must be 1.
Applying this on the given function,
5y^2 - 7y + 1
(1/5) (y^2 - (7/5)y + 1/5)
Therefore, sum of the roots (a + b) and product of the roots (ab), respectively, are:
a + b = 7/5
ab = 1/5
Now, we are asked to find a polynomial with roots 2a/b and 2b/a. What we'll do is get their sum and product:
Product:
2a/b * 2b/a
4ab / ab
= 4
Sum:
2a/b + 2b/a
(2a^2 + 2b^2) / ab
2(a^2 + b^2) / ab
Complete the square inside the parenthesis by adding 2ab, but at the same time subtracting 2(2ab) outside the parenthesis to counter its effect:
[ 2(a^2 + 2ab + b^2) - 4ab ] / ab
[ 2(a + b)^2 - 4ab ] / ab
note that, we have values for a + b and ab from above.
[ 2(7/5)^2 - 4(1/5) ] / (1/5)
= 78/5
Therefore, the polynomial is,
y^2 - (sum)y + (product)
y^2 - (78/5)y + 4
Now try the other problem by using this method.
hope this helps~ `u`
Let beta = b
Because it's easier to type a and b, than copy-pasting alpha and beta symbols.
Anyway, recall that for a quadratic equation in the form
y = ax^2 + bx + c
The sum and product of the roots can be determined as such:
y = x^2 - (sum)x + (product)
Of course, the numerical coefficient of x^2 must be 1.
Applying this on the given function,
5y^2 - 7y + 1
(1/5) (y^2 - (7/5)y + 1/5)
Therefore, sum of the roots (a + b) and product of the roots (ab), respectively, are:
a + b = 7/5
ab = 1/5
Now, we are asked to find a polynomial with roots 2a/b and 2b/a. What we'll do is get their sum and product:
Product:
2a/b * 2b/a
4ab / ab
= 4
Sum:
2a/b + 2b/a
(2a^2 + 2b^2) / ab
2(a^2 + b^2) / ab
Complete the square inside the parenthesis by adding 2ab, but at the same time subtracting 2(2ab) outside the parenthesis to counter its effect:
[ 2(a^2 + 2ab + b^2) - 4ab ] / ab
[ 2(a + b)^2 - 4ab ] / ab
note that, we have values for a + b and ab from above.
[ 2(7/5)^2 - 4(1/5) ] / (1/5)
= 78/5
Therefore, the polynomial is,
y^2 - (sum)y + (product)
y^2 - (78/5)y + 4
Now try the other problem by using this method.
hope this helps~ `u`
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