If alpha and beta are roots of the equation x^2+4x+1=0, what is the value of alpha^2+beta^2?
and also alpha^3 + beta^3
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Equation : x² + 4x + 1 = 0.
Also, it's roots are α and β.
For any equation ax² + bx + c = 0,
sum of it's roots = -b/a
product of it's roots = c/a
So, for the given equation: α + β = -(4)/(1) = -4 and αβ = (1)/(1) = 1.
Using (a+b)² = a² + b² + 2ab
--> a² + b² = (a+b)² - 2ab.
Using the above identity: α² + β² = (α + β)² - 2αβ
--> α² + β² = (-4)² - 2(1)
--> α² + β² = 16 - 2 = 14.
Hence α² + β² = 14.
For α^3 + β^3, using (a + b)^3 = a^3 + b^3 + 3a²b + 3ab²
--> (α + β)^3 - (3a²b + 3ab²) = a^3 + b^3
--> a^3 + b^3 = (a + b)^3 - 3ab(a + b)
Using this, α^3 + β^3 = (α + β)^3 - 3αβ(α + β)
--> α^3 + β^3 = (-4)^3 - 3(1)(-4)
--> α^3 + β^3 = -64 + 12 = -52.
Hence, α^3 + β^3 = -52.
Also, it's roots are α and β.
For any equation ax² + bx + c = 0,
sum of it's roots = -b/a
product of it's roots = c/a
So, for the given equation: α + β = -(4)/(1) = -4 and αβ = (1)/(1) = 1.
Using (a+b)² = a² + b² + 2ab
--> a² + b² = (a+b)² - 2ab.
Using the above identity: α² + β² = (α + β)² - 2αβ
--> α² + β² = (-4)² - 2(1)
--> α² + β² = 16 - 2 = 14.
Hence α² + β² = 14.
For α^3 + β^3, using (a + b)^3 = a^3 + b^3 + 3a²b + 3ab²
--> (α + β)^3 - (3a²b + 3ab²) = a^3 + b^3
--> a^3 + b^3 = (a + b)^3 - 3ab(a + b)
Using this, α^3 + β^3 = (α + β)^3 - 3αβ(α + β)
--> α^3 + β^3 = (-4)^3 - 3(1)(-4)
--> α^3 + β^3 = -64 + 12 = -52.
Hence, α^3 + β^3 = -52.
Abhicris
· 3 years ago
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x^2 + 4x + 1 = 0
Since factoring seems tough, let's complete the square. Move the constant over:
x^2 + 4x = -1
Take the b-term, 4, split in half to get 2, and square to get 4. Add this to both sides:
x^2 + 4x + 4 = 3
Find the squared term:
(x + 2)^2 = 3
Take the square roots:
x + 2 = sqrt(3) and x + 2 = -sqrt(3)
Subtract 2 and we're done:
x = sqrt(3) - 2 and x = -sqrt(3) - 2
So, if alpha = sqrt(3) - 2 and beta = -sqrt(3) - 2, some simple calculations will yield:
alpha^2 + beta^2 = (7 - 4sqrt(3)) + (7 + 4sqrt(3)) = 14
alpha^3 + beta^3 = (15sqrt(3) - 26) + (-15sqrt(3) - 26) = -52 -
Okay, first of all i'm not positive whether i'm completely right, but i'll do my best to help you out. Since it is saying that Beta and alpha are roots of the quadratic then you know that (x + alpha)(x+beta) are your solutions to the quadratic. Now if we foil we end up with x^2 + beta(x) + alpha(x) + (alpha)(beta)
-> x^2 + (beta+alpha)x + (alpha)(beta)
Now for this to be true, we need the coefficients of this quadratic to equal that of the original.. so
x^2 + (beta + alpha)x + (alpha)(beta) = x^2 + 4x + 1
so
beta + alpha = 4
(beta)(alpha)=1
alpha = 4 - beta
(beta)(4-beta) = 1
4beta - beta^2 = 1
or
beta^2 - 4beta +1 = 0
a = 1, b = -4, c = 1
[4 +/- sqrt(12)]/2
[4 +/- 2 sqrt(3)]/2
[2(2 +/- sqrt(3)]/2
beta = 2 +/- sqrt(3)
so imma just take take the positive square root since both of them are actual roots of it.
alpha = 4 -beta -> 4 - (2 + sqrt(3) ) -> 4 - 2 - sqrt(3) -> 2 - sqrt(3)
the initial conditions above where that alpha + beta = 4 and (alpha)(beta) = 1
these conditions are met
So then (2 - sqrt(3))^2 + (2+sqrt(3))^2 = 14
and (2-sqrt(3))^3 + (2+sqrt(3))^3 = 52
Hopefully this is correct, because this is how i would have done it. Cheers! -
x^2 +4x + 1 = 0
Roots:
x = alpha = -3.732050807
x = beta = -0.2679491924
alpha^2 + beta^2
13.92820323 + 0.07179677 = 14
alpha^3 + beta^3
-51.98076209 + -0.019237886 = -52 -
roots/zeros/solutions to the quadratics can be found with the quadratic formula, by factoring, or completing the square.
from inspection, the roots will be irrational and therefore not factorable.
i will find the roots by completing the square:
x^2 + 4x + 1 = 0
(x + 2)^2 = -1 + 4
(x + 2)^2 = 3
x = -2 plus/minus sqrt(3)
thus,
alpa = -2 + sqrt(3)
and
beta = -2 - sqrt(3)
or vice-versa.
either way,
(alpha)^2 + (beta)^2
=
(-2 + sqrt(3))^2 + (-2 - sqrt(3))^2
=
(4 - 4sqrt(3) + 3) + (4 + 4sqrt(3) + 3)
=
14
and
(alpha)^3 + (beta)^3
=
(-2 + sqrt(3))^3 + (-2 - sqrt(3))^3
=
(-8 + 12sqrt(3) - 18 + 3sqrt(3)) + (-8 - 12sqrt(3) - 18 - 3sqrt(3))
=
-52 -
x^2 + 4x + 1 = 0
x = (-4 ± √12) / 2
x = (-4 ± 2√3) / 2
x = -2 ± √3
(-2 + √3)^2 + (-2 - √3)^2 =
4 - 4√3 + 3 + 4 + 4√3 + 3 =
14
(-2 + √3)^3 + (-2 - √3)^3 =
-52 -
∴α and β are roots of the equation x²+4x+1=0.Here,
a=1, b=4, c=1.
∴α + β = −b/a = − 4.
∴αβ = c/a = 1.
∴(α + β)² = (−4)².
⇒α²+β²+2αβ = 16.
⇒α²+β²+1=16.
⇒α²+β²=16− 1.
⇒α²+β²=15.
∴(α + β)³ = (− 4)³.
⇒α³+β³+3αβ(α + β) = −64.
⇒α³+β³+3(1)(−4) = −64.
⇒α³+β³−12 = −64
⇒α³+β³ = −64+12
⇒α³+β³ = −52.
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If alpha and beta are roots of the equation x^2+4x+1=0, what is the value of alpha^2+beta^2?
