Find the following values for the zeros C and B of the polynomial p(x) = x ^ 2 - 5x + 6 without finding the zeros. (1) 1/alpha + 1/beta (2) alpha²+ beta² (3) alpha³+ beta³
Given a quadratic polynomial p(x) = x^2 - 5x + 6, where alpha and beta are the zeros of the polynomial, we can find the following values without explicitly calculating the zeros:
(1) 1/alpha + 1/beta:
Recall that for a quadratic polynomial ax^2 + bx + c, the sum of its zeros (alpha + beta) is given by -b/a, and the product of its zeros (alpha * beta) is given by c/a.
In our case, a = 1, b = -5, and c = 6.
So, alpha + beta = -(-5)/1 = 5, and alpha * beta = 6/1 = 6.
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[tex]\huge\sf\colorbox{lavender}{Answer:-}[/tex]
Given:-
To find:-
[tex] \sf{value \: of \: \frac{1}{ \alpha } + \frac{1}{ \beta }}[/tex]
Solution:-
[tex] \sf{given \: p(x) = x2 - 5x + 6}[/tex]
[tex] \sf{here \: \alpha + \beta - \beta \ \alpha }[/tex]
[tex] \sf{ = - ( - 5) \1}[/tex]
[tex] \sf{5}[/tex]
[tex] \sf{ \alpha a \beta = c \a}[/tex]
[tex] \sf{6 \1}[/tex]
[tex] \sf{6}[/tex]
[tex] \sf{now \: \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{alpha + beta}{a \beta } }[/tex]
[tex] \sf{ \frac{5}{6} }[/tex]
[tex] \sf{value \: of \: the \: expression \: is \: 5 \6}[/tex]
Answer:
Given a quadratic polynomial p(x) = x^2 - 5x + 6, where alpha and beta are the zeros of the polynomial, we can find the following values without explicitly calculating the zeros:
(1) 1/alpha + 1/beta:
Recall that for a quadratic polynomial ax^2 + bx + c, the sum of its zeros (alpha + beta) is given by -b/a, and the product of its zeros (alpha * beta) is given by c/a.
In our case, a = 1, b = -5, and c = 6.
So, alpha + beta = -(-5)/1 = 5, and alpha * beta = 6/1 = 6.
Now, we can find 1/alpha + 1/beta:
1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = 5/6.
(2) alpha² + beta²:
The sum of the squares of the zeros (alpha^2 + beta^2) can be expressed as (alpha + beta)^2 - 2(alpha * beta).
From the previous step, we know that alpha + beta = 5 and alpha * beta = 6.
Now, we can find alpha^2 + beta^2:
alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta) = 5^2 - 2(6) = 25 - 12 = 13.
(3) alpha³ + beta³:
The sum of the cubes of the zeros (alpha^3 + beta^3) can be expressed as (alpha + beta)(alpha^2 - alpha * beta + beta^2).
From the previous steps, we know that alpha + beta = 5, alpha^2 + beta^2 = 13, and alpha * beta = 6.
Now, we can find alpha^3 + beta^3:
alpha^3 + beta^3 = (alpha + beta)(alpha^2 - alpha * beta + beta^2) = 5(13 - 6) = 5(7) = 35.
So, the values are:
(1) 1/alpha + 1/beta = 5/6
(2) alpha² + beta² = 13
(3) alpha³ + beta³ = 35
Step-by-step explanation: