if alpha and beta are the zeros of p(x)=3x^2-7x-6.find a polynomial whose zeros are alpha^2 and beta^2
Thik hai🙃
To find a polynomial whose zeros are the squares of the zeros of the given polynomial, we can use the fact that if α is a zero of p(x), then α^2 is a zero of the polynomial obtained by replacing x with x^2 in p(x).
Let's find the squares of the zeros α and β first:
α = 3x^2 - 7x - 6 = 0
Using the quadratic formula, we can find the value of α:
x = (-(-7) ± √((-7)^2 - 4(3)(-6))) / (2(3))
x = (7 ± √(49 + 72)) / 6
x = (7 ± √121) / 6
x = (7 ± 11) / 6
So, we have two possible values for α: α1 = 3 and α2 = -2/3.
Similarly, let's find the value of β:
β = 3x^2 - 7x - 6 = 0
Using the quadratic formula, we can find the value of β:
x = (-(-7) ± √((-7)^2 - 4(3)(-6))) / (2(3))
x = (7 ± √(49 + 72)) / 6
x = (7 ± √121) / 6
x = (7 ± 11) / 6
So, we have two possible values for β: β1 = 3 and β2 = -2/3.
Now, let's construct the polynomial whose zeros are α^2 and β^2.
The polynomial with zeros α^2 and β^2 can be expressed as:
(x - α^2)(x - β^2) = 0
Substituting the values of α^2 and β^2, we get:
(x - α^2)(x - β^2) = (x - 3^2)(x - (-2/3)^2)
= (x - 9)(x - 4/9)
= x^2 - (9 + 4/9)x + (9 * 4/9)
Expanding and simplifying, we get:
= x^2 - (81/9 + 4/9)x + 36/9
= x^2 - (85/9)x + 4
Therefore, the polynomial whose zeros are α^2 and β^2 is:
p(x) = x^2 - (85/9)x + 4
To find a polynomial whose zeros are the squares of the zeros of another polynomial, you can square the original polynomial's factors.
In this case, the zeros of the polynomial p(x) = 3x^2 - 7x - 6 are alpha and beta.
The square of alpha is alpha^2, and the square of beta is beta^2.
So, we need to find a polynomial whose zeros are alpha^2 and beta^2.
Let's start by finding the factors of p(x). The quadratic equation is given by:
3x^2 - 7x - 6 = 0
To find the zeros (alpha and beta) of this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our case: a = 3, b = -7, and c = -6.
Calculating the roots using the quadratic formula, we get:
alpha = (7 + √(7^2 - 4 * 3 * -6)) / (2 * 3)
beta = (7 - √(7^2 - 4 * 3 * -6)) / (2 * 3)
Simplifying the expressions, we find:
alpha = (7 + √109) / 6
beta = (7 - √109) / 6
Now, to find the polynomial whose zeros are alpha^2 and beta^2, we square the factors:
( x - alpha^2 ) = (x - ( (7 + √109) / 6 )^2 )
( x - beta^2 ) = (x - ( (7 - √109) / 6 )^2 )
Expanding and simplifying the expressions further, we obtain:
( x - alpha^2 ) = (x^2 - (14/6 + √109/6) x + (49/36 + √109/6 + 109/36) )
( x - beta^2 ) = (x^2 - (14/6 - √109/6) x + (49/36 - √109/6 + 109/36) )
Thus, the polynomial whose zeros are alpha^2 and beta^2 is:
p(x) = (x^2 - (14/6 + √109/6) x + (49/36 + √109/6 + 109/36))(x^2 - (14/6 - √109/6) x + (49/36 - √109/6 + 109/36))
(x-3)(3x+2)=0
alpha = 3
beta = -2/3
so
(x-9)(x-4/9) = y
x^2 -9x -4x/9 + 4 = y
y = x^2 - (85/9) x + 4