Solving Quadratic Equations - Pre-Calculus

This is true. The discriminant b2 - 4ac is the part of the quadratic formula that lives inside of a square root function. As you plug in the constants a, b, and c into b2 - 4ac and evaluate, three cases can happen:

b2 - 4ac > 0

b2 - 4ac = 0

b2 - 4ac < 0

In the first case, having a positive number under a square root function will yield a result that is a positive number answer. However, because the quadratic function includes , this scenario yields two real results.

In the middle case (the case of our example), . Going back to the quadratic formula , you can see that when everything under the square root is simply 0, then you get only , which is why you have exactly one real root.

For the final case, if b2 - 4ac < 0, that means you have a negative number under a square root. This means that you will not have any real roots of the equation; however, you will have exactly two imaginary roots of the equation.

This is true. The discriminant b2 - 4ac is the part of the quadratic formula that lives inside of a square root function. As you plug in the constants a, b, and c into b2 - 4ac and evaluate, three cases can happen:

b2 - 4ac > 0

b2 - 4ac = 0

b2 - 4ac < 0

In the first case (the case of our example), having a positive number under a square root function will yield a result that is a positive number answer. However, because the quadratic function includes , this scenario yields two real results.

In the middle case, . Going back to the quadratic formula , you can see that when everything under the square root is simply 0, then you get only , which is why you have exactly one real root.

For the final case, if b2 - 4ac < 0, that means you have a negative number under a square root. This means that you will not have any real roots of the equation; however, you will have exactly two imaginary roots of the equation.

This is false. The discriminant b2 - 4ac is the part of the quadratic formula that lives inside of a square root function. As you plug in the constants a, b, and c into b2 - 4ac and evaluate, three cases can happen:

b2 - 4ac > 0

b2 - 4ac = 0

b2 - 4ac < 0

In the first case, having a positive number under a square root function will yield a result that is a positive number answer. However, because the quadratic function includes , this scenario yields two real results.

In the middle case, . Going back to the quadratic formula , you can see that when everything under the square root is simply 0, then you get only , which is why you have exactly one real root.

For the final case (the case of our example), if b2 - 4ac < 0, that means you have a negative number under a square root. This means that you will not have any real roots of the equation; however, you will have exactly two imaginary roots of the equation.

In the above equation, a = 4, b = 19, and c = -5. Therefore:

b2 - 4ac = (19)2 - 4(4)(-5) = 361 + 80 = 441.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two real roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

which simplifies to

which gives us two answers:

or

These values of x are the two distinct real roots of the given equation.

In the above equation, a = 4, b = 12, and c = 10. Therefore:

b2 - 4ac = (12)2 - 4(4)(10) = 144 - 160 = -16.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two distinct imaginary roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

In other words, our two distinct imaginary roots are and

In the above equation, a = -3, b = 6, and c = -3. Therefore:

b2 - 4ac = (6)2 - 4(-3)(-3) = 36 - 36 = 0.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, there is exactly one real root.

Finally, we use the quadratic function to find these exact root. The quadratic formula is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

which simplifies to

which gives us one answer: x = 1

This value of x is the one distinct real root of the given equation.

For any quadtratic equation of the form , the quadratic formula is

Plugging in our given values we have:

In the above equation, a = 1, b = 5, and c = 4. Therefore:

b2 - 4ac = (5)2 - 4(1)(4) = 25 - 16 = 9.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two real roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

which simplifies to

which gives us two answers:

x = -1 or x = -4

These values of x are the two distinct real roots of the given equation.

In the above equation, a = -1, b = 3, and c = -3. Therefore:

b2 - 4ac = (3)2 - 4(-1)(-3) = 9 - 12 = -3.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two distinct imaginary roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

Because , this simplifies to . In other words, our two distinct imaginary roots are and

In the above equation, a = 1, b = 2, and c = 10. Therefore:

b2 - 4ac = (2)2 - 4(1)(10) = 4 - 40 = -36.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, we have two distinct imaginary roots.

Finally, we use the quadratic function to find these exact roots. The quadratic function is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

Because , this simplifies to . We can further simplify this to . In other words, our two distinct imaginary roots are and .

In the above equation, a = 1, b = 8, and c = 16. Therefore:

b2 - 4ac = (8)2 - 4(1)(16) = 64 - 64 = 0.

When the discriminant is greater than 0, there are two distinct real roots. When the discriminant is equal to 0, there is exactly one real root. When the discriminant is less than zero, there are no real roots, but there are exactly two distinct imaginary roots. In this case, there is exactly one real root.

Finally, we use the quadratic function to find these exact root. The quadratic formula is:

Plugging in our values of a, b, and c, we get:

This simplifies to:

which simplifies to

This value of x is the one distinct real root of the given equation.

To find the roots of an equation in the form , you use the quadratic formula

.

In our case, we have .

This gives us which simplifies to

When solving a quadratic equation, the first thing to look for is whether or not it can be factored, as this is most often the easiest and fastest method if the quadratic can in fact be factored. We can see that each of the terms in the given equation have a common factor of 3, so it will be easier to factor the quadratic if we first factor out the 3:

Now we're left with a polynomial where we need to find two numbers whose product is -28 and whose sum is -3. Thinking about the factors of 28, we can see that 4 and 7 will yield -3 if 7 is negative and 4 is positive, so we now have our factorization:

Factorize and set this equation equal to zero.

The answer is one of the possible choices.

There are two solutions; .

We proceed as follows.

Add to both sides.

Take the square root of both sides, remember to introduce plus/minus on the right side since you are introducing a square root into your work.

Add to both sides.

Use either the quadratic formula or factoring to solve the quadratic equation.

Using factoring, we want to find which factors of six when multiplied with the factors of two and then added together result in negative one.

Using the quadratic formula,

let

To solve this equation, use trial and error to factor it. Since the leading coefficient is , there is only one way to get , so that is helpful reminder. Once it's properly factored, you get: . Then, set both of those expressions equal to to get your roots: .

If the roots of the equation are at x= -4 and x=3, then we can work backwards to see what equation those roots were derived from. If we factored a quadratic equation and obtained the given solutions, it would mean the factored form looked something like:

Because this is the form that would yield the solutions x= -4 and x=3. If we work backwards and multiply the factors back together, we get the following quadratic equation:

We can make a quadratic polynomial with by mutiplying the linear polynomials they are roots of, and multiplying them out.

Start

Distribute the negative sign

FOIL the two polynomials. This means multiply the firsts, then the outers, followed by the inners and lastly, the last terms.

Simplify

These two points tell us that the quadratic function has zeros at , and at .

These correspond to the linear expressions , and .

Expand their product and you arrive at the correct answer.

If the quadratic is opening up the coefficient infront of the squared term will be positive. Thus we get:

.

If the quadratic is opening down it would pass through the same two points but have the equation:

.

Since only is seen in the answer choices, it is the correct answer.