The quadratic polynomial p(x) = ax^2 + bx + c has positive integer coefficients a, b and c that each have a value from 1 to 9, inclusive. To eight decimal places, the solutions to p(x) = 0 are x = -0.18858048 and x = -0.58919729. What is the vaue of p(10)?

\(\phantom{x = -0.19098301 and x = -1.30901699.}\)

Guest Jul 9, 2022

#1**-1 **

Suppose that the roots are r, s

P(x) = ax^2 + bx + c = (x - r)(x-s)

Since we know what r and s is, we know that P(10) is equal to (10 + 0.18859048)(10 + 0.58919729)

Yikes, this doens't seem like the right path.

We may want to find the coefficients instead of the exact value of each of these decimals. If we use vieta's formula, you DONT need a calculator(I will show this magic to you :) )

We know that the product of these two roots is equal to c/a and that it's a very tiny value. Any value greater than 1/9 will surely not equal the product of these two numbers. Therefore, c = 1, a = 9..

By vieta's, b/a is approximately 0.77(You don't hae to expliclity sum up ALL those values).. Therefore, since a = 9, b is equal to 7.

Therefore, our polynomial is:

P(x) = 9x^2 + 7x + 1

We will plug in 10 and this gives us \(\boxed{971}\)

Voldemort Jul 9, 2022