QuestionFind the values of $m$ such that $2x - 2m + (x + m)^{2} > 0$ for all real $x$ .

Studdy Solution

STEP 1

Assumptions1. The expression $x-m+(x+m)^{}$ is given. . We need to find the range of values of $m$ for which this expression is positive for all real values of $x$ .

STEP 2

First, let's simplify the given expression. $2x-2m+(x+m)^{2} =2x -2m + x^2 +2xm + m^2$

STEP 3

Rearrange the terms and combine like terms.
$2x-2m+(x+m)^{2} = x^2 +2xm -2m +2x + m^2$

STEP 4

Further simplify the expression.
$2x-2m+(x+m)^{2} = x^2 +2x(m+1) + m^2 -2m$

STEP 5

Now, we can see that the expression is a quadratic equation in the form of $ax^2 + bx + c$ . For a quadratic equation to be positive for all real values of $x$ , the discriminant $b^2 -4ac$ must be less than0.

STEP 6

The coefficients of the quadratic equation are $a=1$ , $b=2(m+1)$ , and $c=m^2-2m$ . Let's substitute these values into the discriminant inequality.
$(2(m+1))^2 -4(1)(m^2-2m) <0$

STEP 7

implify the inequality.
$4(m^2+2m+1) -4m^2 +m <0$

STEP 8

Further simplify the inequality.
$4m^2 +8m +4 -4m^2 +8m <0$

STEP 9

Combine like terms.
$16m +4 <$

STEP 10

olve the inequality for $m$ .
$16m < -4$

STEP 11

Divide both sides of the inequality by16.
$m < -\frac{}{4}$ So, the range of values for $m$ for which $x-m+(x+m)^{}$ is positive for all real values of $x$ is $m < -\frac{}{4}$ .

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