+610 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 $a+b$?
+128408 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
+2440 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.
