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1.) Find all pairs of real numbers $(x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. If you find more than one pair, then list your pairs in order by increasing $x$ value, separated by commas. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(-3,9),(2,4)" (without the quotation marks).

 

2.)The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

gueesstt  Apr 22, 2018
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5+0 Answers

 #1
avatar+86613 
+1

2.)The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i

 

The sum of the roots  =   -a / 1

So

(2 - 3i) + (2 + 3i)  = - a /1

4  =  -a

-4  = a

 

The product of the roots will  =  b /1

So

(2 - 3i) (2 + 3i)  =   b

4 + 9  = 13 = b

 

So....a + b  =   -4 + 13   =   9

 

 

cool cool cool

CPhill  Apr 22, 2018
 #4
avatar+604 
+2

Thank you!!

gueesstt  Apr 22, 2018
 #2
avatar+1961 
+2

The first equation involves knowledge on how to solve system of equations. There is a chance that there will be two solutions here because the second equation \(x^2+y^2=28\) is a degree-two polynomial. I think it would be easiest to utilize the substitution method here.

 

\(x+y=6\) I will solve for x, but you could just as easily solve for y.
\(x=6-y\)  
   

 

Since I know that x=6-y, I can substitute this information into the second equation \(x^2+y^2=28\).

 

\(x^2+y^2=28\) Substitute x for 6-y and solve the equation.
\((6-y)^2+y^2=28\) Let's expand the left hand side of the equation.
\(36-12y+y^2+y^2=28\) Let's do some rearranging and combining of like terms.
\(2y^2-12y+36=28\) Subtract 28 from both sides.
\(2y^2-12y+8=0\) Every term on the left hand side has a factor of 2, so let's fact that out!
\(2(y^2-6y+4)=0\) Unfortunately, the trinomial inside the parentheses is not factorable, so we will have to use alternate methods (like the quadratic formula) to figure out the roots of this equation.
\(a=1;b-6;c=4\\ y_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(4)}}{2(1)}\) The only thing to do now is simplify.
\(y_{1,2}=\frac{6\pm\sqrt{20}}{2}\) Let's simplify the radical completely.
\(y_{1,2}=\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5} \)  
\(y_1=3+\sqrt{5}\\ y_2=3-\sqrt{5}\) We now have to figure out what the corresponding x-values are. Let's use the first equation for that (\(x+y=6\)).

 

\(x_1+y_1=6\) Substitute the value for y_1 and solve for the remaining unknown.
\(x_1+3+\sqrt{5}=6\) Subtract 3
\(x_1+\sqrt{5}=3\) Subtract \(\sqrt{5}\)
\(x_1=3-\sqrt{5}\) Now, let's figure out the next x-value.
\(x_2+y_2=6\)  
\(x_2+3-\sqrt{5}=6\) Now, do the same process.
\(x_2-\sqrt{5}=3\)  
\(x_2=3+\sqrt{5}\)  
   

 

Therefore, the two solutions for this system are \((3-\sqrt{5},3+\sqrt{5}),(3+\sqrt{5},3-\sqrt{5})\). This is just part one.

TheXSquaredFactor  Apr 22, 2018
 #3
avatar+604 
+2

Thank you!!

gueesstt  Apr 22, 2018
 #5
avatar+1961 
+2

No problemo!

TheXSquaredFactor  Apr 22, 2018

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