Question $k \in \left(-\infty, \frac{11}{7}\right) \cup \left(\frac{11}{7}, \infty\right)$

Studdy Solution

STEP 1

Assumptions1. The equation is -7w + (k-1) = -4w^ . We are looking for all values of $k$ for which the equation has two real solutions3. The equation is a quadratic equation in the form $ax^ + bx + c =0$ , where $a$ , $b$ , and $c$ are constants and $x$ is the variable. In this case, $w$ is the variable, $-4$ is $a$ , $-7$ is $b$ , and $(k-1)$ is $c$ .

STEP 2

For a quadratic equation to have two real solutions, the discriminant must be greater than zero. The discriminant is given by $b^2 -4ac$ .
$Discriminant = b^2 -4ac$

STEP 3

Plug in the values for $a$ , $b$ , and $c$ into the discriminant formula.
$Discriminant = (-7)^2 -(-)(k-1)$

STEP 4

implify the discriminant.
$Discriminant =49 +16(k-1)$

STEP 5

For the equation to have two real solutions, the discriminant must be greater than zero.
$49 +16(k-1) >0$

STEP 6

implify the inequality.
$16k -16 +49 >0$

STEP 7

Combine like terms.
$16k +33 >0$

STEP 8

Isolate $k$ on one side of the inequality.
$16k > -33$

STEP 9

Divide both sides of the inequality by $16$ to solve for $k$ .
$k > -\frac{33}{16}$ All values of $k$ greater than $-\frac{33}{16}$ will result in the equation having two real solutions.

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