How do you show that one of the roots of x2+ax+b=0 is a, if and only if b=0?
Answers
There's some mistake in the question.
If and only if b = 0, then one of the roots of the equation x² + ax + b = 0 is -a, not a.
Consider the given polynomial p(x) = x² + ax + b = 0.
First we have to assume that b = 0.
Therefore,
⇒ x² + ax + b = 0
⇒ x² + ax = 0
⇒ x(x + a) = 0
From this, we get that the roots of the polynomial are 0 and -a.
Hence Proved!
But actually we have to prove that one of the roots of p(x) is -a if and only if b = 0.
We've proved that x = -a if b = 0. Now we're going to prove that b = 0 if x = -a, means vice versa!
Assume that one of the roots of a quadratic polynomial is -a.
Thus, one of the factors of the quadratic polynomial will be (x + a).
We have found earlier that the other root of the polynomial x² + ax + b = 0 is 0, thus the other factor will be x. So (x + a) has to be multiplied by x.
Thus,
⇒ x(x + a) = 0
⇒ x² + ax = 0
Here a quadratic polynomial x² + ax = 0 is obtained.
The polynomial given in the question is x² + ax + b = 0.
As both results in 0,
⇒ x² + ax = x² + ax + b
⇒ 0 = b
Hence it's proved!!!
Answer:
Step-by-step explanation:
Take b=0 prove that -a is a root of x2 +ax +b =0
Given b=0
x2 +ax =0
x[x+a] =0
x=0 or x+a=0
0 and -a are roots ⇒ a is a root
Take -a is a root of x2 +ax +b =0 prove that b=0
-a is a root of x2 +ax +b =0
[-a]2 +a[-a] +b=0
a2 - a2 + b=0
b=0
hence proved