If the zeroes of the quadratic polynomial x^2+(a+1)x+bare 1& -3 find a&b
well, geez, just plug in the values. You have
(x-1)(x+3)
= x^2 + 2x - 3 = x^2 + (a+1)x + b
Now it should be clear what a and b are if the two polynomials are identical.
Most likely the topic is the Sum and Product of roots of a quadratic
for ax^2 + bx + c = 0
sum of roots = -b/a
product of roots = c/a
given:
sum of roots = -3+1 = -2
product of roots = -3
then -(a+1) = -2
a = 1
b/1 = -3
b = -3
oh, yeah. Good call.
To find the values of "a" and "b" in the quadratic polynomial x^2 + (a+1)x + b, when the zeroes are 1 and -3, you can use the fact that the sum of the zeroes of a quadratic polynomial is equal to the negation of the coefficient of x divided by the coefficient of x^2, and the product of the zeroes is equal to the constant term divided by the coefficient of x^2.
Given that the zeroes are 1 and -3, we can use these relationships to set up two equations and solve for "a" and "b":
Equation 1: Sum of zeroes = -b/a = 1 + (-3) = -2
Equation 2: Product of zeroes = b/a = 1 * (-3) = -3
Now, we have a system of two equations:
-b/a = -2
b/a = -3
To eliminate the fraction, we can multiply both equations by "a":
Equation 1: -b = -2a
Equation 2: b = -3a
Now, we can substitute Equation 2 into Equation 1 to solve for "a":
-(-3a) = -2a
3a = -2a
5a = 0
a = 0
Once we have determined the value of "a" as 0, we can substitute it back into one of the original equations to solve for "b". Let's use Equation 2:
b = -3(0)
b = 0
Therefore, the values of "a" and "b" are both 0 for the given quadratic polynomial.