if alpha and beta are the zeros of the polynomial 2x^2-4x+5 then find the values of
(i)alpha^2+beta^2
(ii)1/alpha^2+1/beta^2
(iii)(r)alpha/beta+(r)beta/alpha
(iv)alpha^-1+beta^-1
alpha+beta=-b/a
=--4/2
=2
alpha^2+beta^2=(alpha+beta)^2
=2^2
=4
(ii)take reciprocal(1/2)^2=1/4
(iv)a^-1=1/a similarly 1/2
tried my best to answer hope its correct:)
your pal
mathew
To find the values for the given expressions, we need to use the properties and relationships between zeros of a polynomial.
Given that alpha and beta are the zeros of the polynomial 2x^2 - 4x + 5, we can use the following relationships:
1) Sum of zeros: alpha + beta = -b/a
2) Product of zeros: alpha * beta = c/a
In this case, a = 2, b = -4, and c = 5.
Let's proceed step-by-step:
(i) To find alpha^2 + beta^2:
Since (alpha + beta)^2 - 2alpha*beta = alpha^2 + beta^2, we need to find the values of (alpha + beta)^2 and 2alpha*beta first.
Using the sum and product relationships:
(alpha + beta)^2 = (-(-4) / 2)^2 = (4 / 2)^2 = 2^2 = 4
2alpha * beta = 2 * (5 / 2) = 5
Therefore, alpha^2 + beta^2 = 4 - 5 = -1.
(ii) To find 1/alpha^2 + 1/beta^2:
Using the relationship:
1/alpha^2 + 1/beta^2 = (alpha^2 + beta^2) / (alpha * beta)^2
We already calculated alpha^2 + beta^2 as -1, and (alpha * beta) = (5 / 2). So,
1/alpha^2 + 1/beta^2 = (-1) / (5/2)^2 = (-1) / (25/4) = (-4) / 25.
(iii) To find (r)alpha/beta + (r)beta/alpha:
Using the relationship:
(r)alpha/beta + (r)beta/alpha = [(r * alpha^2) + (r * beta^2)] / (alpha * beta)
We can rewrite the expression as:
(r * alpha^2 + r * beta^2) / (alpha * beta)
Since we know alpha^2 + beta^2 = -1 and (alpha * beta) = (5 / 2), we have:
(r * alpha^2 + r * beta^2) / (alpha * beta) = (r * (-1) + r * (-1)) / (5 / 2) = -2r / (5 / 2) = -4r / 5.
(iv) To find alpha^-1 + beta^-1:
Using the relationships:
alpha^-1 + beta^-1 = (alpha + beta) / (alpha * beta)
We have already calculated (alpha + beta) as -(-4) / 2 = 4 / 2 = 2, and (alpha * beta) as 5 / 2.
Therefore, alpha^-1 + beta^-1 = 2 / (5 / 2) = 4 / 5.
So, the values for the given expressions are:
(i) alpha^2 + beta^2 = -1
(ii) 1/alpha^2 + 1/beta^2 = -4/25
(iii) (r)alpha/beta + (r)beta/alpha = -4r/5
(iv) alpha^-1 + beta^-1 = 4/5
To find the values of the given expressions, we need to first find the values of α and β, which are the zeros of the polynomial 2x^2 - 4x + 5.
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, we have the quadratic equation 2x^2 - 4x + 5 = 0. Comparing it to the general form, we can see that a = 2, b = -4, and c = 5.
Applying the quadratic formula, we have:
α = (-(-4) ± √((-4)^2 - 4 * 2 * 5))/(2 * 2)
= (4 ± √(16 - 40))/4
= (4 ± √(-24))/4
Since the discriminant (√(b^2 - 4ac)) is negative, the roots will be complex conjugates. Let's use the imaginary unit i, where i^2 = -1, to simplify the expression:
α = (4 ± 2i√6)/4
= 1 ± i√6/2
= 1/2 ± i√6/2
Therefore, α = 1/2 + i√6/2 and β = 1/2 - i√6/2 (complex conjugate of α).
Now, let's solve each part of the question:
(i) To find α^2 + β^2, we can directly substitute the values of α and β:
α^2 + β^2 = (1/2 + i√6/2)^2 + (1/2 - i√6/2)^2
= 1/4 + i^2(√6)^2/4 + 2i^2(1/2)(√6)/4 + 1/4 - i^2(√6)^2/4 - 2i^2(1/2)(√6)/4
= 1/4 - 6/4 + i√6/2 + 1/4 - 6/4 - i√6/2
= 2/4 - 12/4
= -10/4
= -5/2
Therefore, α^2 + β^2 = -5/2.
(ii) To find 1/α^2 + 1/β^2, we can calculate the reciprocal of α and β first:
1/α = 1/(1/2 + i√6/2)
= 1/(1/2) * (2/(1/2 + i√6/2))
= 2/(1 + i√6)
Similarly, 1/β = 2/(1 - i√6)
Now, substituting the values:
1/α^2 + 1/β^2 = (2/(1 + i√6))^2 + (2/(1 - i√6))^2
= 4/(1 + 2i√6 + (i√6)^2) + 4/(1 - 2i√6 + (i√6)^2)
= 4/(1 - 6 + 2i√6) + 4/(1 - 6 - 2i√6)
= 4/(-5 + 2i√6) + 4/(-5 - 2i√6)
= (4 (-5 - 2i√6) + 4 (-5 + 2i√6))/((-5 + 2i√6) (-5 - 2i√6))
= (-20 - 8i√6 - 20 + 8i√6)/((-5)^2 - (2i√6)^2)
= (-40)/(-25 + 24)
= -40/-1
= 40
Therefore, 1/α^2 + 1/β^2 = 40.
(iii) To find (r)α/β + (r)β/α, we nee
The general solution to the equation
ax^2+bx+c=0
is, using the quadratic formula:
α=(-b+√(b²-4ac))/2a, and
β=(-b-√(b²-4ac))/2a
From there, expand α²+β² to get (b²-2ac)/a²
I will leave parts (ii) to (iv) for you as practice.