Show that x² + (a? - 2)x - 2a = 0 and x - 3x + 2 = 0 have exactly one common root for all a € R.
Find the minimum value of :
Answers
Given: The equations x² + (a^2 - 2)x - 2a^2 = 0 and x - 3x + 2 = 0
To find: Show that these equations have exactly one common root for all a € R.
Solution:
- Now we have given the equations as :
x² + (a^2 - 2)x - 2a^2 = 0 and x - 3x + 2 = 0
- Consider x - 3x + 2 = 0 , we have:
x^2 -x - 2x + 2 = 0
x( x - 1 ) - 2( x - 1 ) = 0
(x - 2)(x - 1) = 0
x = 2 , 1
- Now putting these values in x² + (a^2 - 2)x - 2a^2 = 0, we get:
- First putting 2, we get:
2^2 + (a^2 - 2)2 - 2a^2 = 0
4 + 2a^2 - 4 - 2a^2 = 0
0 = 0
- Now putting 1, we get:
1^2 + (a^2 - 2)1 - 2a^2 = 0
1 + a^2 - 2 - 2a^2 = 0
-1 - a^2 = 0
a^2 = -1
a = √-1 .............not possible.
- So the common root is x = 2.
Answer:
So these equations has exactly one common root for all a € R and the root is x = 2.
Answer: 2 is the common root of this equations.
Step-by-step explanation:
Now we have given the equations as :
x² + (a^2 - 2)x - 2a^2 = 0 and x - 3x + 2 = 0
Consider x - 3x + 2 = 0 , we have:
x^2 -x - 2x + 2 = 0
x( x - 1 ) - 2( x - 1 ) = 0
(x - 2)(x - 1) = 0
x = 2 , 1
Now putting these values in x² + (a^2 - 2)x - 2a^2 = 0, we get:
First putting 2, we get:
2^2 + (a^2 - 2)2 - 2a^2 = 0
4 + 2a^2 - 4 - 2a^2 = 0
0 = 0
Now putting 1, we get:
1^2 + (a^2 - 2)1 - 2a^2 = 0
1 + a^2 - 2 - 2a^2 = 0
-1 - a^2 = 0
a^2 = -1
a = √-1 .............not possible.
So the common root is 2