#7:
Recall that [tex]x^2+2x+1[/tex] is a perfect square, since it factors into [tex](x+1)^2[/tex] nicely. We can add 1 to both sides of the equation and then factor, in this case, the [tex]a^2+2a[/tex] and the 1.
[tex]a^2+2a+1-50=5+1[/tex]
[tex](a+1)^2-50=6[/tex]
[tex](a+1)^2-56=0\\[/tex] (subtract 6 both sides)
#8:
Instead of remembering perfect squares, we can find what c value makes a quadratic perfect: c = [tex](\frac{b}{2})^2[/tex].
So in this case, [tex]x^2-12x+36[/tex] makes a perfect square, and we can verify by quickly factoring it to be [tex](x-6)^2[/tex]. So, we add 36 on both sides and do the same thing as we did for 7!
[tex]x^2-12x+36+15=-9+36[/tex]
[tex](x-6)^2+15=27[/tex]
[tex](x-6)^2-12=0[/tex] (subtract 27 both sides)
Now try 9 and 10 on your own!
Let me know if you get stuck or have any other questions!
Answer:
Step-by-step explanation:
Step 1: Add 50 to both sides to leave room for completing the square.
a^2 + 2a -50 +50 = 5 +50 gets you a^2 +2a = 55 (Notice that I left room after a^2 = 2a.
Step 2: To complete the square use (b/2)^2 in the space left for this process. In this case, b=2 so (2/2)^2 = 1^2. Leave it as 1^2 which I will explain.
Step 3: if you add 1^2 which is 1 to the left side, you must add it to the right side i.e. 55+1 = 56. So you should have now:
a^2 +2a + 1^2 = 56. When you complete the square, the reason you want to leave the 1^2 instead of 1 is that you can now show the factorization by taking the a^2 and the 1^2 and making it (a+1)^2 = 56
Step 4: Take the square root of both sides:
[tex]\sqrt(a+1)^{2} = \sqrt56[/tex] which gives you a+1 = [tex]\sqrt56[/tex] which can be simplified by subtracting 1 from both sides to
a= 1 [tex]\frac{+}{-}[/tex] [tex]\sqrt{56}\\[/tex] where [tex]\sqrt{56} = \sqrt{4} x \sqrt{14}[/tex] which is 2[tex]\sqrt{14}[/tex]
So the final answer is 1 [tex]\frac{+}{-}[/tex] 2[tex]\sqrt{14}[/tex]
Each of the other problems is done the same way 832 963 2185 for explanation
