Question

algebra 2
m3 quiz 2
solve each equation by factoring. must show all work for credit.

1. $x^{2}+8x=-16$
2. $n^{2}=-49 + 14n$

solve each equation by taking square roots. must show all work for credit.

1. $2b^{2}+6=-36$
2. $4x^{2}+10=-70$

solve each equation by completing the square. must show all work for credit.

1. $x^{2}-14x - 70=-3$
2. $p^{2}-4p - 30=-9$

Explanation:

Step1: Rearrange equation 1

Rewrite $x^{2}+8x = - 16$ as $x^{2}+8x + 16=0$.

Step2: Factor equation 1

Using the perfect - square formula $(a + b)^2=a^{2}+2ab + b^{2}$, where $a = x$ and $b = 4$, we get $(x + 4)^{2}=0$. Then $x+4 = 0$, so $x=-4$.

Step3: Rearrange equation 2

Rewrite $n^{2}=-49 + 14n$ as $n^{2}-14n + 49 = 0$.

Step4: Factor equation 2

Using the perfect - square formula $(a - b)^2=a^{2}-2ab + b^{2}$, where $a = n$ and $b = 7$, we get $(n - 7)^{2}=0$. Then $n-7 = 0$, so $n = 7$.

Step5: Rearrange equation 3

Rewrite $2b^{2}+6=-36$ as $2b^{2}=-42$, then $b^{2}=-21$. Since the square of a real number cannot be negative, there are no real - valued solutions for $b$.

Step6: Rearrange equation 4

Rewrite $4x^{2}+10=-70$ as $4x^{2}=-80$, then $x^{2}=-20$. Since the square of a real number cannot be negative, there are no real - valued solutions for $x$.

Step7: Complete the square for equation 5

Rewrite $x^{2}-14x-70=-3$ as $x^{2}-14x=67$. Add $(\frac{-14}{2})^{2}=49$ to both sides: $x^{2}-14x + 49=67 + 49$, which is $(x - 7)^{2}=116$. Then $x-7=\pm\sqrt{116}=\pm2\sqrt{29}$, so $x = 7\pm2\sqrt{29}$.

Step8: Complete the square for equation 6

Rewrite $p^{2}-4p-30=-9$ as $p^{2}-4p=21$. Add $(\frac{-4}{2})^{2}=4$ to both sides: $p^{2}-4p + 4=21 + 4$, which is $(p - 2)^{2}=25$. Then $p-2=\pm5$, so $p=2\pm5$. The solutions are $p = 7$ and $p=-3$.

Answer:

1. $x=-4$
2. $n = 7$
3. No real - valued solutions
4. No real - valued solutions
5. $x = 7\pm2\sqrt{29}$
6. $p = 7,p=-3$