+2446 This one requires some thinking. We know that a quadratic equation has 2 zeroes. One is located at (x,0), and the other is located at (y,0).
Now, what do we know about the relationship of these two numbers? Well, we know that the sum of both zeroes is -15.
| x+y=−15 | Let's solve for y here. |
| y=−15−x | |
Since y=-15-x, we know that a zero is located at (−15−x,0). We also know that, when multiplied, both zeroes yield -54. Knowing this information, we can generate a quadratic equation.
x(−15−x)=−54
This quadratic equation has everything that we are looking for. Now, solve for the zeroes of this quadratic.
| x(−15−x)=−54 | Distribute the x into the binomial in the parentheses. | ||
| −15x−x2=−54 | Add 54 to both sides. | ||
| −x2−15x+54=0 | Divide by -1. This is not strictly necessary, but it does make the x^2-term positive, which could make methods of solving easier. | ||
| x2+15x−54=0 | Factor this by finding two numbers that multiply to -54 and add to 15. These happen to be -3 and 18. | ||
| (x−3)(x+18)=0 | By the zero product theorem, set both products equal to zero. | ||
| Add the constant in both equations and move it move to the right hand side. | ||
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These zeroes definitely fit the criteria.
+2446 Evaluating the expression of 38178−16
| 38178−16 | First, convert the fraction in the denominator into an improper fraction. |
| 38178−16=388∗1+78−16=38158−16 | Multiply by the reciprocal of the denominator to eliminate it. |
| 815815∗38158−16 | Doing this isn't actually changing the value of the fraction because we are just multiplying by 1. |
| 815∗38−16 | Before beginning the multiplication, we can drastically simplify the numbers in both fractions by identifying the GCF of opposite numerators and denominators. In this example, 8 and 8 have a GCF of 8. 3 and 15 have a GCF if 3. |
| 15−16 | Convert 1/5 and 1/6 into fractions with a common denominator. |
| 630−530 | Now subtract the numerators while maintaining the denominator. |
| 130 | |
