1.Write a quadratic equation in the variable x having the given numbers as solutions. Type the equation in stadard form ax^2+bx+c=0
Solution 7, only solution
2.The width of a rectangle is less than the length. The area is 4ft^3. Find the length and the width.
1. If a quadratic has only one real solution, then there has to be a double factor
So in this case
(x-7)(x-7) = 0
expand if needed
2. Question makes no sense.
area cannot be in cubic units
1. To find a quadratic equation that has a single solution, we need to use the formula: (x - r)^2 = 0, where r is the given solution.
In this case, the given solution is 7. So, (x - 7)^2 = 0.
To write it in standard form, we need to expand the equation: x^2 - 14x + 49 = 0.
Therefore, the quadratic equation with 7 as its only solution is: x^2 - 14x + 49 = 0.
2. Let's assume that the width of the rectangle is w and the length is l.
According to the given information, width < length, so we can write the inequality as w < l.
The area of the rectangle is given as 4ft^3, which can be expressed as the product of width and length: w * l = 4.
We know that width is less than length, but we don't have enough information to determine the exact values of w and l. However, we can try different combinations of w and l that satisfy the area equation.
Let's consider a few possibilities:
- w = 1, l = 4: 1 * 4 = 4 (satisfies the condition)
- w = 2, l = 2: 2 * 2 = 4 (satisfies the condition)
- w = 0.5, l = 8: 0.5 * 8 = 4 (satisfies the condition)
As we can see, there are multiple valid combinations of width and length that satisfy the given conditions.