Question

What is the lowest value of x2 + 4x + 2?
Ans : 2
How?
Please explain in details.​

Answer 1

We have: y=x2+2x+1

Let’s factorise this expression using the middle-term break:

=x2+x+x+1

=x(x+1)+1(x+1)

=(x+1)(x+1)

=(x+1)2

Now, this expression is squared, so it always produces zero or a positive result, i.e. (x+1)2≥0 .

⇒(x+1)2≥0

⇒x+1≥0

⇒x≥−1

The minimum value of x is −1 . Let’s substitute this into the equation to find the minimum value of y :

⇒y=(−1)2+2(−1)+1

=1−2+1

=2−2

=0

Therefore, the minimum value of y=x2+2x+1 is 0 .

Answer 2

SOLUTION

TO DETERMINE

The the lowest value of the expression

$\sf{ {x}^{2} + 4x + 2}$

EVALUATION

Here the given expression is

$\sf{ {x}^{2} + 4x + 2}$

Now

$\sf{ {x}^{2} + 4x + 2}$

$\sf{ = {x}^{2} + 2.x.2 + {2}^{2} - {2}^{2} + 2}$

$\sf{ = {(x + 2)}^{2} - 4 + 2}$

$\sf{ = {(x + 2)}^{2} - 2}$

Now minimum value of ( x + 2 )² is 0 [ when x = - 2 ]

Hence the required lowest value of the expression

= 0 - 2

= - 2

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