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# Help

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1. Let f(x)=ax^2 +bx+a,where a and b are constants and a≠0. If one of the roots of the equation f(x)=0 is x=4, what is the other root?

2. Find the sum of the coefficients of the product (x^50+x^49+x^48+···+x^2+x+1)(x^50−x^49+x^48 −···+x^2 −x+1).

(In the second factor, the coefficients alternate between 1 and −1.)

Thank you.

Jun 15, 2024

#1
+1690
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Problem 1:

We can leverage the concept of sum and product of roots in quadratic equations.

Since x = 4 is a root, we know that f(4) = 0. This translates to:

16a + 4b + a = 0

Combining like terms, we get: 17a + 4b = 0 (Equation 1)

We are also given that f(x) = ax^2 + bx + a. Since a ≠ 0, we can rewrite f(x) by factoring out a:

f(x) = a(x^2 + bx/a + 1/a)

We know that one root is x = 4. This means that (x - 4) is a factor of f(x). Let's rewrite f(x) using this information:

f(x) = a(x - 4)(x + r) for some constant r (since the product of the roots is cr)

Equating both ways of expressing f(x), we get:

a(x^2 + bx/a + 1/a) = a(x - 4)(x + r)

Expanding both sides:

ax^2 + bx + a = ax^2 - (4a + r)x + 4ar

Equating the coefficients of like terms on both sides, we get:

b = - (4a + r) (Equation 2)

Now we have two independent equations (Equation 1 and Equation 2) with unknowns a, b, and r. We can use the fact that one root is x = 4 (reflected in Equation 1) to solve for the other root.

Since we're looking for the other root, let's try to eliminate a and b from the equations and solve for r.

From Equation 1: b = -17a/4

Substitute this into Equation 2:

-17a/4 = - (4a + r)

Solve for r:

17a/4 = 4a + r

r = 17a/4 - 4a

r = a/4

Now that we know r (product of roots), we can find the other root itself. Since the product of roots (cr) is also equal to f(4) (which is 0 in this case), we have:

(4)(other root) = 0

Since a ≠ 0, this implies the other root must be:

Other root = 0

In conclusion, the other root of the equation f(x) = 0 is x = 0.

Jun 15, 2024
#2
+1690
0

Problem 2

Note that both factors share some common terms. Let's rewrite them to group these terms together:

(x^50 + x^48 + ... + x^2 + 1) + (x^49 - x^47 + ... + x - 1)

According to the sum of a finite geometric series formula, the sum of each individual sequence is:

(x^50 + x^48 + ... + x^2 + 1) = (x^(50)) * (1 / (1 - x^2)) = x^(50) / (1 - x^2)

(x^49 - x^47 + ... + x - 1) = (x^(49)) * (1 / (1 + x^2)) = x^(49) / (1 + x^2)

Now we can find the product:

Product = (x^(50) / (1 - x^2)) * (x^(49) / (1 + x^2))

Since we're only interested in the sum of the coefficients, we can ignore the constant terms in the denominators (they won't affect the sum). This simplifies the product to:

Product ≈ x^(50 + 49) // Ignoring constant terms in denominators

The sum of the coefficients in this product is simply the coefficient of the term with x raised to the highest power, which is x^(99). The coefficient of this term is 1.

In conclusion, the sum of the coefficients of the product is 1.

Jun 15, 2024
#3
+129690
+1

1.  ax^2 + bx + a

Product of the roots =  a/a   = 1

So....call the second root, m

And

m * 4  = 1

m =  1/4 = the other root

Jun 15, 2024
#4
+22
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Yall were both so helpful thank you.

Jun 15, 2024