Plug in y = 1. Then solve for x.

2x + y³ + xy² + y = x

2x + 1 + x + 1 = x

3x + 2 = x

2x = -2

x = -1

-intercepts will occur when . This yields the equation

We need to use the quadratic formula where ,  and .

Plugging in our values:

Simplifying: 

Simplifying:

Simplifying:

Finally:

The area of a rectangle is the product of its length by its width, which we know to be equal to  in our problem. We also know that the length is equal to , where  represents the width of the land. Therefore, we can write the following equation:

Distributing the  outside the parentheses, we get:

Subtracting  from each side of the equation, we get:

We get a quadratic equation, and since there is no factor of  and  that adds up to , we use the quadratic formula to solve this equation.

We can first calculate the discriminant (i.e. the part under the square root)

We replace that value in the quadratic formula, solving both the positive version of the formula (on the left) and the negative version of the formula (on the right):

Breaking down the square root:

We can pull two of the twos out of the square root and place a  outside of it:

We can then multiply the  and the :

At this point, we can reduce the equations, since each of the component parts of their right sides has a factor of :

Since width is a positive value, the answer is:

 The width of the piece of land is approximately .

Solve by factoring.  First get everything into the form Ax² + Bx + C = 0:

x² + 4x - 5 = 0

Then factor: (x + 5) (x - 1) = 0

Solve each multiple separately for 0:

X + 5 = 0; x = -5

x - 1 = 0; x = 1

Therefore, x is either -5 or 1

Begin by multiplying both sides by (x – 1):

x² – x = x – 1

Solve as a quadratic equation: x² – 2x + 1 = 0

Factor the left: (x – 1)(x – 1) = 0

Therefore,  x = 1.

However, notice that in the original equation, a value of 1 for x would place a 0 in the denominator.  Therefore, there is no solution.

Set up two equations from the given information:

 and 

Substitute into the second equation:

Multiply through by .

Then divide by the coefficient of 2 to simplify your work:

Then since you have a quadratic setup, move the  term to the other side (via subtraction from both sides) to set everything equal to 0:

As you look for numbers that multiply to positive 120 and add to -22 so you can factor the quadratic, you might recognize that -12 and -10 fit the bill. This makes your factorization:

This makes the possible solutions 10 and 12. Since 12 does not appear in the choices,  is the only possible correct answer.

We need to input 1.5 into our function, then we need to input "a" into our function and set these results equal.

f(a) = f(1.5)

f(a) = -(1.5)² +6(1.5) -5

f(a) = -2.25 + 9 - 5

f(a) = 1.75

-a² + 6a -5 = 1.75

Multiply both sides by 4, so that we can work with only whole numbers coefficients.

-4a² + 24a - 20 = 7

Subtract 7 from both sides.

-4a² + 24a - 27 = 0

Multiply both sides by negative one, just to make more positive coefficients, which are usually easier to work with.

4a² - 24a + 27 = 0

In order to factor this, we need to mutiply the outer coefficients, which gives us 4(27) = 108. We need to think of two numbers that multiply to give us 108, but add to give us -24. These two numbers are -6 and -18. Now we rewrite the equation as:

4a² - 6a -18a + 27 = 0

We can now group the first two terms and the last two terms, and then we can factor.

(4a² - 6a )+(-18a + 27) = 0

2a(2a-3) + -9(2a - 3) = 0

(2a-9)(2a-3) = 0

This means that 2a - 9 =0, or 2a - 3 = 0.

2a - 9 = 0

2a = 9

a = 9/2 = 4.5

2a - 3 = 0

a = 3/2 = 1.5

So a can be either 1.5 or 4.5.

The only answer choice available that could be a is 4.5.

Quadratic equations generally have two answers.  We add 5 to both sides and then divide by 2 to get the quadratic expression on one side of the equation: (x + 1)² = 16.   By taking the square root of both sides we get x + 1 = –4 or x + 1 = 4.  Then we subtract 1 from both sides to get x = –5 or x = 3.

Generally, quadratic equations have two answers.

First, the equations must be put in standard form: 3x² + 10x + 3 = 0

Second, try to factor the quadratic; however, if that is not possible use the quadratic formula.

Third, check the answer by plugging the answers back into the original equation.

Begin by getting our equation into the form Ax² + BX + C = 0:

3x² – 11x + 10 = 0

Now, if you factor the left, you can find the answer. Begin by considering the two groups.  They will have to begin respectively with 3 and 1 as coefficients for your x value.  Likewise, looking at the last element, you can tell that both will have to have a + or –, since the C coefficient is positive.  Finally, since the B coefficient is negative, we know that it will have to be –. We know therefore:

(3x – ?)(x – ?)

The potential factors of 10 are: 10, 1; 1, 10; 2, 5; 5, 2

5 and 2 work:

(3x – 5)(x – 2) = 0 because you can FOIL (3x – 5)(x – 2) back into 3x² – 11x + 10.

Now, the trick remaining is to set each of the factors equal to 0 because if either group is 0, the whole equation will be 0:

3x – 5 = 0 → 3x = 5 → x = 5/3

x – 2 = 0 → x = 2

Therefore, x is either 5 / 3 or 2. The former is presented as an answer.