Consider the polynomial f(x)=17x4+21x3+60x2+Ax+B. Suppose that for every root λ of f(x)=0, 1λ is also a root of f(x)=0. What is the value of A+B?
what is 1λ ?
A missing symbol somewhere?
1/λ?
1-λ?
1+λ?
actually it's 1/λ
haha this is a feautured problem on Brilliants :')
Yes ... ! :P LOL
To find the value of A+B, we need to make use of the given information about the roots of the polynomial.
Let's start by assuming that λ is a root of f(x) = 0. According to the information given, we know that 1/λ is also a root of f(x) = 0.
Based on this, we can write the following equation:
f(x) = (x - λ)(x - 1/λ)(g(x))
Where g(x) is a quadratic polynomial with the remaining two roots of f(x) = 0.
Expanding this equation, we get:
f(x) = (x^2 - xλ - λ/x + 1)(g(x))
Now, let's compare this expanded equation with the given polynomial f(x) = 17x^4 + 21x^3 + 60x^2 + Ax + B.
Comparing the coefficients of corresponding terms on both sides of the equation, we can derive the following relationships:
x^4 term: 1 = 17 (since we have no other similar term on the right-hand side)
x^3 term: -λ - 1/λ = 21
x^2 term: 1 - λ(1/λ) + 1 = 60
x term: -λ - 1/λ = A
constant term: 1(1/λ) = B
From the equations above, we can solve for λ and find the corresponding values of A and B.
Let's solve these equations step by step:
1. From the first equation, 1 = 17, we can see that 1 is not equal to 17. Hence, there are no real solutions for λ in this case.
2. As a result, we can deduce that λ must be a complex number, specifically a non-real number since there are no real solutions to the equation.
Let's denote the non-real value of λ as i*ϕ, where ϕ is a real number.
Now, substituting i*ϕ into the second equation, we get:
-i*ϕ - 1/(i*ϕ) = 21
To simplify this equation, we can multiply both sides by i*ϕ to eliminate the denominators:
-i*ϕ * (i*ϕ) - 1 = 21 * (i*ϕ)
ϕ^2 - 1 = 21i*ϕ
For this equation to hold true, the real and imaginary parts on both sides of the equation must be equal.
Equating the real parts, we get:
ϕ^2 - 1 = 0
Solving this quadratic equation, we find two real solutions: ϕ = 1 and ϕ = -1.
To find the values of A and B, we can substitute each solution of ϕ back into the equations for A and B:
A = -i*ϕ - 1/(i*ϕ)
A = -i*1 - 1/(i*1)
A = -i - 1/(-i)
A = -i - (i/(-i*i))
A = -i + i/i^2
A = -i + i/-1
A = -i - i
A = -2i
B = 1/(i*ϕ)
B = 1/(i*1)
B = 1/i
B = -i
To find the value of A + B, we can add the values we obtained:
A + B = -2i + (-i)
A + B = -2i - i
A + B = -3i
Therefore, the value of A + B is -3i.