Quadratic Equation - GRE Quantitative Reasoning

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.

36x2 - 12x - 15 = 0

Factor the equation:

(6x + 3)(6x - 5) = 0

Set each side equal to zero

6x + 3 = 0

x = -3/6 = -1/2

6x – 5 = 0

x = 5/6

64x2 + 24x - 10 = 0

Factor the equation:

(8x + 5)(8x – 2) = 0

Set each side equal to zero

(8x + 5) = 0

x = -5/8

(8x – 2) = 0

x = 2/8 = 1/4

The roots of an equation are the points at which the function equals zero. We can set each function equal to zero and determine which functions have one root, and which does not.

Another piece of information will help. If a quadratic function has one root, then it must be a perfect square. This is because a quadratic function that is a perfect square can be written in the form (x – a)2. If we set (x – a)2 = 0 in order to find the root, we see that a is the only value that solves the equation, and thus a is the only root. Additionally, a quadratic equation is a perfect square if it can be written in the form a2x2 + 2abx + b2 = (ax + b)2.

Let's examine the choice f(x) = 4x2 – 4x+1. To find the roots, we set f(x) = 0.

4x2 – 4x+1 = 0

We notice that 4x2 - 4x + 1 is a perfect square, since we could write it as (2x – 1)2. Thus, this equation has only one root, and it can't be the answer.

If we look at f(x) = x2 –2x + 1, we see that x2 – 2x + 1 is also a perfect square, because it could be written as (x – 1)2. This function also has a single root.

Next, we examine f(x) = (1/4)x2 + x + 1. Let us set f(x) = (1/4)x2 + x + 1 = 0.

(1/4)x2 + x + 1 = 0

We can multiply both sides by four to get rid of the fraction.

x2 + 4x + 4 = 0

(x + 2)2 = 0

This function is also a perfect square and has a single root.

Now consider the choice f(x) = (–1/9)x2 + 6x – 81.

f(x) = (–1/9)x2 + 6x – 81 = 0

Multiply both sides by –9.

x2 – 54x + 729 = 0

(x – 27)2 = 0.

Finally, let's look at f(x) = 9x2 – 6x + 4. This CANNOT be written as a perfect square, because it is not in the form a2x2 + 2abx + b2 = (ax + b)2. It might be tempting to think that 9x2 - 6x + 4 = (3x - 2)2, but it does NOT, because (3x – 2)2 = 9x2 – 12x + 4. Therefore, because 9x2 – 6x + 4 is not a perfect square, it doesn't have exactly one root.

x2 – x = 72. Solve for x using the quadratic formula and x = 9 and –8. Only 9 satisfies the restrictions.

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

(x + 2)(x + 3) = 0

x = –2 or x = –3

The roots of a function are the x intercepts of the function. Whenever a function passes through a point on the x-axis, the value of the function is zero. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values of x.

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

Because the coefficient in front of the is not equal to 1, we need to multiply this coefficient by the constant, which is –4. When we mutiply 2 and –4, we get –8. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of the x term). Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We will now rewrite –7x as –8x + x.

We will then group the first two terms and the last two terms.

We will next factor out a 2_x_ from the first two terms.

Thus, when factored, the original equation becomes (2_x_ + 1)(x – 4) = 0.

We now set each factor equal to zero and solve for x.

Subtract 1 from both sides.

2_x_ = –1

Divide both sides by 2.

Now, we set x – 4 equal to 0.

x – 4 = 0

Add 4 to both sides.

x = 4

The roots of f(x) occur where x = .

The answer is therefore .

We can factor the quadratic equation into .

Then we can see that .

Therefore, becomes and .

To solve, begin by factoring the equation. We know that in our two factors, one will begin with and one will begin with . Fiddling with different factors of (we are looking for two numbers that, when multiplied by and separately, will add to ), we come to the following:

(If you are unsure, double check by expanding the equation to match the original)

Now, set the each factor equal to 0:

Do the same for the second factor:

Therefore, our two values of are and .

Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.

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.

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.

We can use the quadratic formula to find the solutions to quadratic equations in the form ax2 + bx + c = 0. The quadratic formula is given below.

In order for the formula to give us real solutions, the value under the square root, b2 – 4ac, must be greater than or equal to zero. Otherwise, the formula will require us to find the square root of a negative number, which gives an imaginary (non-real) result.

In other words, we need to look at each equation and determine the value of b2 – 4ac. If the value of b2 – 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x2 – 4x + 5 = 0 and determine the value of b2 – 4ac.

b2 – 4ac = (–4)2 – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b2 – 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x2 – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b2 – 4ac, and thus these other equations will have real solutions.

We are told that f(x) = x2 - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k2 – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k2 – 4k +2 = 6 – k

Add k to both sides.

k2 – 3k + 2 = 6

Then subtract 6 from both sides.

k2 – 3k – 4 = 0

Factor the quadratic equation. We must think of two numbers that multiply to give us -4 and that add to give us -3. These two numbers are –4 and 1.

(k – 4)(k + 1) = 0

Now we set each factor equal to 0 and solve for k.

k – 4 = 0

k = 4

k + 1 = 0

k = –1

The two possible values of k are -1 and 4. The question asks us to find their sum, which is 3.

The answer is 3.

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.

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.

The equation we are asked to solve is 16x – 10(4)x + 16 = 0.

Equations of this type can often be "transformed" into other equations, such as linear or quadratic equations, if we rewrite some of the terms.

First, we can notice that 16 = 42. Thus, we can write 16x as (42)x or as (4x)2.

Now, the equation is (4x)2 – 10(4)x + 16 = 0

Let's introduce the variable u, and set it equal 4x. The advantage of this is that it allows us to "transform" the original equation into a quadratic equation.

u2 – 10u + 16 = 0

This is an equation with which we are much more familiar. In order to solve it, we need to factor it and set each factor equal to zero. In order to factor it, we must think of two numbers that multiply to give us 16 and add to give us –10. These two numbers are –8 and –2. Thus, we can factor u2 – 10u + 16 = 0 as follows:

(u – 8)(u – 2) = 0

Next, we set each factor equal to 0.

u – 8 = 0

Add 8.

u = 8

u – 2 = 0

Add 2.

u = 2.

Thus, u must equal 2 or 8. However, we want to find x, not u. Since we defined u as equal to 4x, the equations become:

4x = 2 or 4x = 8

Let's solve 4x = 2 first. We can rewrite 4x as (22)x = 22x, so that the bases are the same.

22x = 2 = 21

2x = 1

x = 1/2

Finally, we will solve 4x = 8. Once again, let's write 4x as 22x. We can also write 8 as 23.

22x = 23

2x = 3

x = 3/2

The original question asks us to find the sum of the values of x that solve the equation. Because x can be 1/2 or 3/2, the sum of 1/2 and 3/2 is 2.

The answer is 2.