Solved on Nov 15, 2023

k(,117)(117,)k \in \left(-\infty, \frac{11}{7}\right) \cup \left(\frac{11}{7}, \infty\right)

STEP 1

Assumptions1. The equation is -7w + (k-1) = -4w^ . We are looking for all values of kk for which the equation has two real solutions3. The equation is a quadratic equation in the form ax+bx+c=0ax^ + bx + c =0, where aa, bb, and cc are constants and xx is the variable. In this case, ww is the variable, 4-4 is aa, 7-7 is bb, and (k1)(k-1) is cc.

STEP 2

For a quadratic equation to have two real solutions, the discriminant must be greater than zero. The discriminant is given by b24acb^2 -4ac.
Discriminant=b24acDiscriminant = b^2 -4ac

STEP 3

Plug in the values for aa, bb, and cc into the discriminant formula.
Discriminant=(7)2()(k1)Discriminant = (-7)^2 -(-)(k-1)

STEP 4

implify the discriminant.
Discriminant=49+16(k1)Discriminant =49 +16(k-1)

STEP 5

For the equation to have two real solutions, the discriminant must be greater than zero.
49+16(k1)>049 +16(k-1) >0

STEP 6

implify the inequality.
16k16+49>016k -16 +49 >0

STEP 7

Combine like terms.
16k+33>016k +33 >0

STEP 8

Isolate kk on one side of the inequality.
16k>3316k > -33

STEP 9

Divide both sides of the inequality by 1616 to solve for kk.
k>3316k > -\frac{33}{16}All values of kk greater than 3316-\frac{33}{16} will result in the equation having two real solutions.

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