Which one is real?
How many real solutions does the following equation have?
x 2 + 4 x + 2 0 + 2 x 2 + 8 x + 1 2 = 6
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2 solutions
It has two roots but they are equal.
@Utsav Banerjee Exactly the same!!!
Oh, i got using the terrible substititution, x 2 + 4 x = y .
Well can't it be |16|+|22|=6.
Let the L H S be f ( x ) . Then
f ( x ) = x 2 + 4 x + 2 0 + 2 x 2 + 8 x + 1 2 = ( x + 2 ) 2 + 1 6 + 2 ( x + 2 ) 2 + 4 Since ( x + 2 ) 2 ≥ 0
⟹ min ( f ( x ) ) = f ( − 2 ) = 1 6 + 4 = 6 = R H S . For all other x = − 2 , f ( x ) > 6 . Therefore, there is only 1 solution.
x 2 + 4 x + 2 0 = ( x + 2 ) 2 + 1 6 ≥ 1 6
2 x 2 + 8 x + 1 2 = 2 ( x + 2 ) 2 + 4 ≥ 4
Therefore, x 2 + 4 x + 2 0 + 2 x 2 + 8 x + 1 2 ≥ 4 + 2 = 6
with equality occurring when
x 2 + 4 x + 2 0 = 1 6 and 2 x 2 + 8 x + 1 2 = 4
⇒ x + 2 = 0
⇒ x = − 2
So, x = − 2 is the only solution