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.

x² – 3x – 18 = 0 can be factored to (x – 6)(x + 3) = 0

Therefore, x can be either 6 or –3.

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.

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem.  However, it is possible to factor this if you are careful.  Factored, the equation can be rewritten as:

Now, either one of the groups on the left could be  and the whole equation would be .  Therefore, you set up each as a separate equation and solve for :

OR

The sum of these values is:

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

Now, while you could use the quadratic formula to solve this problem, the easiest way to work this question is to factor the left side of the equation.  This gives you:

Now, either one of the groups on the left could be  and the whole equation would be .  Therefore, you set up each as a separate equation and solve for :

OR

Since both of your answers are less than , quantity B is larger than quantity A.

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.

Define the variables to be x = first multiple of three and x + 3 = the next consecutive multiple of 3.

Knowing the product of these two numbers is 54 we get the equation x(x + 3) = 54. To solve this quadratic equation we need to multiply it out and set it to zero then factor it. So x² + 3x – 54 = 0 becomes (x + 9)(x – 6) = 0.  Solving for x we get x = –9 or x = 6 and only the positive number is correct.  So the two numbers are 6 and 9 and their sum is 15.

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.