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-1
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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 $a+b$?

Apr 22, 2018

#1
+101088
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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

Apr 22, 2018
#4
+603
+1

Thank you!!

gueesstt  Apr 22, 2018
#2
+2339
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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.

Apr 22, 2018
#3
+603
+1

Thank you!!

gueesstt  Apr 22, 2018
#5
+2339
+1

No problemo!

TheXSquaredFactor  Apr 22, 2018