Final Answer:
The original equation with these coefficients would be x² + x - 2 = 0, the values of b and c are b = 1 and c = -2.
Step-by-step explanation:
To find the coefficients b and c for the quadratic equation x² + bx + c = 0, where b and c are also the roots of that equation, we will make use of Vieta's formulas.
These formulas relate the coefficients of a polynomial to sums and products of its roots.
For a quadratic equation of the form x² + px + q = 0, Vieta's formulas tell us that:
- The sum of the roots (-b/a) is equal to -p (where a is the leading coefficient, which is 1 in this case since the equation is x² + bx + c).
- The product of the roots (c/a) is equal to q (where a is the leading coefficient, which is 1 in this case as well).
But in our problem, the sum of the roots is equal to b, and the product of the roots is equal to c. We are given that the roots are equal to the coefficients, which means:
- Let the roots of the equation be b and c.
- Therefore, b + c = -b (from Vieta's formulas, sum of the roots).
- Also, bc = c (from Vieta's formulas, product of the roots).
We have two equations here:
1. b + c = -b
2. bc = c
Let's solve equation 1 for c:
b + c = -b
c = -b - b
c = -2b
Now let's substitute c from equation 1 into equation 2:
b(-2b) = -2b
bc = c
-2b^2 = -2b
Since we are looking for a non-zero solution (if b = 0, then c must also be 0, and that would make the quadratic trivial), we can divide both sides of this equation by -2b, assuming b is not equal to 0:
b = 1
Now, we substitute b = 1 back into the equation c = -2b to find c:
c = -2(1)
c = -2
Therefore, the values of b and c for which the coefficients of the quadratic equation are also its roots are b = 1 and c = -2.
Complete question:
The coefficients b and c in the equation x²+bx+c = 0 are also the roots of that equation. Find b and c.