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The equation

2 x^{2} + q x + q - 1 = 0

has exactly two real solutions.

Find the range of possible values of $q$ .

Use the table below to find your answer based on your solution set.

Give your answer to $3$ significant figures.

Hint

Consider the discriminant.

Hint

b^{2} - 4 a c > 0

Solution

Exactly two real solution means that the discriminant is positive

\begin{aligned} b^{2} - 4 a c & > 0 \\ q^{2} - 4 (2) (q - 1) & > 0 \\ q^{2} - 8 q + 8 & > 0 \end{aligned}

The roots are $q = 4 \pm 2 \sqrt{2}$ and we sketch $q^{2} - 8 q + 8$ :

rectangle

We see that $q^{2} - 8 q + 8 > 0$ when

q < 4 - 2 \sqrt{2} \cup q > 4 + 2 \sqrt{2}