one root of the quadratic equation ax2 + bx + c = 0 is 0, then –
Answers
Answer:
Let one root be α
Then the other root is α
n
So,product of roots =
a
c
∴(α)(α
n
)=
a
c
∴α
n+1
=
a
c
∴α=(
a
c
)
n+1
1
...(1)
sum of roots =−
a
b
∴α+α
n
=−
a
b
Substituting the value of α from equation (1), we get
∴(
a
c
)
n+1
1
+(
a
c
)
n+1
n
=−
a
b
∴a
n+1
n
c
n+1
1
+a
n+1
1
c
n+1
n
=−b
Step-by-step explanation:
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Answer :
Other root = -b/a
Note :
- The values of the variable which satisfy any equation are called its roots or solutions .
Solution :
Here ,
The given quadratic equation is ;
ax² + bx + c = 0 .
Also ,
It is given that , x = 0 is a root of the given quadratic equation . Thus , x = 0 must satisfy the given quadratic equation .
Now ,
Putting x = 0 , the given quadratic equation ,
We get ;
=> a•0² + b•0 + c = 0
=> 0 + 0 + c = 0
=> c = 0
Now ,
Putting c = 0 , the given quadratic equation will reduce to ;
=> ax² + bx + c = 0
=> ax² + bx + 0 = 0
=> ax² + bx = 0
Now ,
=> ax² + bx = 0
=> x(ax + b) = 0
=> x = 0 , -b/a
Clearly ,
The other root of the given quadratic equation is (-b/a) .
Alternative method :
Note :
★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;
• Sum of roots , (α + ß) = -b/a
• Product of roots , (αß) = c/a
Here ,
It is given that , one of the root of the quadratic equation ax² + bx + c = 0 is 0 .
Thus ,
Let α = 0
Also ,
We know that , the sum of roots of the quadratic equation ax² + bx + c = 0 is given as (-b/a) .
Thus ,
=> α + ß = -b/a
=> 0 + ß = -b/a
=> ß = -b/a