+2446 It does not appear as if Cphill answered the first question of for what value of c makes \(2x^2+8x+c=0\) have only 1 unique solution.
At first, I was not sure how to approach it, but then I remembered something called the discriminant. The discriminant, in a quadratic equation \(ax^2+bx+c\) is \(b^2-4ac\). The discriminant can tell you a few things about the number of solutions in a quadratic.
1) If \(b^2-4ac>0\), then there are 2 unique solutions
2) If \(b^2-4ac<0\), then there are no real solutions
3) If \(b^2-4ac=0\), then there is 1 unique solution
Based on the rules I had just described, we should definitely use condition #3 because we want the quadratic to have 1 unique solution.
| \(b^2-4ac=0\) | Plug in the values we know. |
| \((-8)^2-4(2)(c)=0\) | Let's simplify the right hand side. |
| \(64-8c=0\) | Subtract 64 from both sides of the equation. |
| \(-8c=-64\) | Divide by -8 on both sides. |
| \(c=8\) | |
Therefore, \(c=8\) is the only value for c such that there is only one unique solution.
+2446 I actually do not like the method of completing the square because I believe it to be quite inefficient, but I'll use it, if you insist.
Using the method of completing the square, I will solve for x in the equation \(-2x^2+5x+1=0\).
| \(-2x^2+5x+1=0\) | In a quadratic equation in the form, \(ax^2+bx+c=0\) we must move c to the other side; in other words, subtract 1 on both sides of the equation. | ||
| \(-2x^2+5x=-1\) | Before we can proceed, we must make the coefficient of the quadratic term 1. At the moment, it is -2. Therefore, we must divide by -2 on both sides. | ||
| \(x^2-\frac{5}{2}x=\frac{1}{2}\) | Now, the next part is probably the hardest to understand. Our goal in the next step is make \(x^2-\frac{5}{2}x\) a perfect-square trinomial by adding a value. In order to do that, we must add \(\left(\frac{b}{2}\right)^2\) where b is the coefficient of the linear term. | ||
| \(x^2-\frac{5}{2}x+\left(\frac{b}{2}\right)^2=\frac{1}{2}+\left(\frac{b}{2}\right)^2\) | I added (b/2)^2 to both sides because whatever you do one side, you must do to the other in order to keep the equations balanced. Now, plug in the appropriate value of b, (-5/2) in this case. | ||
| \(x^2-\frac{5}{2}x+\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{1}{2}+\left(\frac{\frac{-5}{2}}{2}\right)^2\) | Now, I will simplify within the parentheses. | ||
| \(\frac{\left(\frac{-5}{2}\right)}{2}=\frac{-5}{2}\div\frac{2}{1}=\frac{-5}{2}*\frac{1}{2}=\frac{-5}{4}\) | Reinsert this. | ||
| \(x^2-\frac{5}{2}x+\left(\frac{-5}{4}\right)^2=\frac{1}{2}+\left(\frac{-5}{4}\right)^2\) | Distribute the exponent to both the numerator and denominator. | ||
| \(x^2-\frac{5}{2}x+\frac{(-5)^2}{4^2}=\frac{1}{2}+\frac{(-5)^2}{4^2}\) | Simplify both the numerator and the denominator of both fractions. | ||
| \(x^2-\frac{5}{2}x+\frac{25}{16}=\frac{1}{2}+\frac{25}{16}\) | Let's simplify the right hand side of the equation. In this case, this requires converting 1/2 into a fraction with the denominator of 16. | ||
| \(\frac{1}{2}+\frac{25}{16}=\frac{8}{16}+\frac{25}{16}=\frac{33}{16}\) | |||
| \(x^2-\frac{5}{2}x+\frac{25}{16}=\frac{33}{16}\) | By adding (b/2)^2 to both sides of the equation, we have created a perfect-square trinomial on the left-hand side. It may not be obvious because of the fractions, but it may help to know that the trinomial, when manipulated correctly, will always become in the form of \(\left(x+\frac{b}{2}\right)^2\). | ||
| \(\left(x-\frac{5}{4}\right)^2=\frac{33}{16}\) | Take the square root of both sides. Remember that the square root of both sides results in both the absolute value. | ||
| \(\left|x-\frac{5}{4}\right|=\sqrt{\frac{33}{16}}\) | Let's simplify the right hand side of the equation by distributing the square root to both the numerator and the denominator. | ||
| \(\left|x-\frac{5}{4}\right|={\frac{\sqrt{33}}{\sqrt{16}}}\) | The square root of 16 is 4, so let's simplify that. | ||
| \(\left|x-\frac{5}{4}\right|=\frac{\sqrt{33}}{4}\) | Now, solve for both positive and negative answer. | ||
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+2446 To find the vertical asymptotes, we must figure out when the denominator of \(\frac{5x^2-9}{3x^2+5x+2}\) equals 0. This will give us the vertical asymptotes. Let's do that!
| \(3x^2+5x+2=0\) | I will use the AC method to factor out this. What number multiplies to get 6 and adds to get 5? 3 and 2, of course! Break up the b-term. | ||
| \(3x^2+3x+2x+2=0\) | Let's solve this by grouping. I'll use parentheses to make it easier to follow, I hope. | ||
| \((3x^2+3x)+(2x+2)=0\) | Factor out the GCF of both expressions inside of the parentheses. | ||
| \(3x(x+1)+2(x+1)=0\) | Now, use the reverse-distributive-property to combine 3x and 2. In other words, \(ac+bc=(a+b)c\) | ||
| \((3x+2)(x+1)=0\) | Set both factors equal to 0 and solve both independently. | ||
| Subtract the constant term from both sides of the equation. | ||
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x=-(2/3) and x=-1 are the location of both vertical asymptotes on the graph.
Now, add both answers together to get the result of \(a+b\).
\(-1-\frac{2}{3}=\frac{-3}{3}-\frac{2}{3}=-\frac{5}{3}=-1.\overline{66}\)
.+2446 Coincidentally, trigonometry was unnecessary in this problem (although use whatever solving method you'd like!).
The height of the rectangle (we know it is a rectangle because it a quadrilateral with 4 right angles) \(FLAX\) is of equal length to the height of\(\triangle ZAX\).
Let's draw in \(\overline{ZB}\) such that the segment is a perpendicular bisector of \(\overline{FL}\) and \(B\) lies on the midpoint of \(\overline{XA}\).
Inserting the altitude of an isosceles (or equilateral, in this case) triangle has 2 lesser-known properties.
1) The altitude bisects the intersected angle.
2) The altitude bisects the intersected segment.
Put all of this information together \(m\angle AZB=30^{\circ}\hspace{1mm}\text{and}\hspace{1mm}m\angle ZBA=90^{\circ}\hspace{1mm}\text{and}\hspace{1mm} m\angle BAZ=60^{\circ}\). \(AB=5\hspace{1mm}\text{and}\hspace{1mm} AZ=10\)
It's a 30-60-90 triangle. Yay! By definition, the ratio of the side lengths is \(1:\sqrt{3}:2\). Knowing this, we can calculate the length of the altitude.
| \(\frac{ZB}{\sqrt{3}}=\frac{5}{1}\) | Multiply by the square root of 3 on both sides of the equation. |
| \(ZB=5\sqrt{3}\) | |
It has already been established that ZB is the height of the rectangle. Therefore, multiply the length and the width together to get the total area.
\(A=10*5\sqrt{3}=50\sqrt{3}units^2\)
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