Question

solve the following quadratic equations. remember, you are to use completing the square to solve these equations. each answer must be in set notation (i.e. use the fancy brackets).
show your work!!!!

1. $k^2 + 12k - 34 = 0$
2. $a^2 + 8a + 62 = 6$
3. $n^2 + 2n - 75 = 7$
4. $x^2 - 8x - 45 = 0$
5. $b^2 - 8b + 79 = 3$

Explanation:

Step1: Isolate constant term (Q7)

$a^2 + 8a = 6 - 62$
$a^2 + 8a = -56$

Step2: Complete the square (Q7)

Take half of 8, square it: $(\frac{8}{2})^2=16$. Add to both sides.
$a^2 + 8a + 16 = -56 + 16$
$(a+4)^2 = -40$

Step3: Solve for $a$ (Q7)

Take square roots of both sides.
$a+4 = \pm\sqrt{-40} = \pm2i\sqrt{10}$
$a = -4 \pm 2i\sqrt{10}$

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Step1: Isolate constant term (Q8)

$n^2 + 2n = 7 + 75$
$n^2 + 2n = 82$

Step2: Complete the square (Q8)

Take half of 2, square it: $(\frac{2}{2})^2=1$. Add to both sides.
$n^2 + 2n + 1 = 82 + 1$
$(n+1)^2 = 83$

Step3: Solve for $n$ (Q8)

Take square roots of both sides.
$n+1 = \pm\sqrt{83}$
$n = -1 \pm \sqrt{83}$

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Step1: Isolate constant term (Q9)

$x^2 - 8x = 45$

Step2: Complete the square (Q9)

Take half of -8, square it: $(\frac{-8}{2})^2=16$. Add to both sides.
$x^2 - 8x + 16 = 45 + 16$
$(x-4)^2 = 61$

Step3: Solve for $x$ (Q9)

Take square roots of both sides.
$x-4 = \pm\sqrt{61}$
$x = 4 \pm \sqrt{61}$

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Step1: Isolate constant term (Q10)

$b^2 - 8b = 3 - 79$
$b^2 - 8b = -76$

Step2: Complete the square (Q10)

Take half of -8, square it: $(\frac{-8}{2})^2=16$. Add to both sides.
$b^2 - 8b + 16 = -76 + 16$
$(b-4)^2 = -60$

Step3: Solve for $b$ (Q10)

Take square roots of both sides.
$b-4 = \pm\sqrt{-60} = \pm2i\sqrt{15}$
$b = 4 \pm 2i\sqrt{15}$

Answer:

1. $\{-4 - 2i\sqrt{10}, -4 + 2i\sqrt{10}\}$
2. $\{-1 - \sqrt{83}, -1 + \sqrt{83}\}$
3. $\{4 - \sqrt{61}, 4 + \sqrt{61}\}$
4. $\{4 - 2i\sqrt{15}, 4 + 2i\sqrt{15}\}$