Week 8 In Pre-Calculus 11: Determining the Nature of x with the Discriminant

I decided to do this because you can skip some steps if you find out the discriminant early ex. if discriminant is negative.

In a quadratic equation, we can determine the nature of x with the help of the discriminant. The discriminant refers to the radical in the quadratic formula,

 

The nature of x can be determined with the following guidelines.

  • If the radical is positive, there are 2 solutions (real & unequal).
  • If the radical is 0, there is 1 solution (real & equal).
  • If the radical is negative, there is no solution (non-real).

 

Let’s predict how many solutions a quadratic equation will have and verify.

By this we can see that there will be 1 solution.

 

Example 2:

As we can see the radical is negative, meaning there’s no real solution. There isn’t really a point in trying to verify since the radical won’t magically change.

 

Example 3:

The radical is positive, there will be 2 solutions.

If the discriminant isn’t a perfect square, it is usually best to leave the radical as an absolute value. You will still have 2 answers hence the +/- sign.

Week 7 In Pre-Calculus 11 – Solving Quadratic Equations: Quadratic Formula

I chose to do this because it looks like it will be very useful down the line.

So far, we have learned two ways to solve quadratic equations, by factoring and completing the perfect square. This time, we learned another method.

All in all, the formula is not that complicated, you just need to have the equation in general quadratic equation form so you can put the values in the formula. The general form of a quadratic equation is .

The quadratic formula uses this form, with it having the formula

Let us solve an example,

You might notice that it doesn’t factor normally very easily, so the best approach would be using the quadratic formula (or completing the square).

By looking at this, we can see that a=3, b=7, and c=1.

Putting this into the formula:

Now, we do algebra as normal. Simplify to

Since the radical can’t be simplified to a mixed and there is no common factors, this would be the final exact value answer.

Week 6 in Pre-Calculus 11 – Solving Radical Equations Using Factoring

I chose this because it is more challenging to me, so explaining it will be more beneficial.

To solve for x, we must first get rid of the radical by squaring both sides, but first we have to isolate the radical, rewrite the equation as

Now we can square both sides, remember that 2x+3 is a binomial, meaning squaring it will give you a trinomial.

To be able to solve for x, one side of the equation needs to equal 0, so we subtract 19x and 6 from both sides

Now we can factor it normally to get

If the answer is 0, we need to make x equal something that when multiplied by the other binomial makes a 0. This much is simple since multiplying anything by 0 gives 0, so we just need to make x equal something that results in one binomial equating to 0.

In this case, there are 2 possible answers. 1 (1 times 1 minus 1 is 0), and 3/4 (4 times 3/4 minus 3 is 0).

However, the answers won’t always be correct. We have to check the answer by using the value x to see if the equation is true.

For x=1, is true.

 

For x=3/4,

is also true, making both solutions correct

Week 5 in Pre-Calculus 11 – Factoring with Substitution

I chose this because it seems like a useful and easy method to do on most factoring questions that have multiple binomials.

When I’m talking about substituting, I’m talking about replacing a binomial with a variable, for example (3a-4)(a+2) turns into AB, where A is (3a-4) and B is (a+2).

Let’s look at how we can use this to factor.

When we look at the binomials, we notice that we can substitute the binomials for variables to be able to simplify. Let (3a-4) = A, and (a+2) = B

Now the expression has been simplified to

With the expression like this, it is easier to see how we can write this as a product of binomials,

We’re not done yet though, because we have to remember that the variables were substituted. Substituting back gives us:

We can distribute to get ,

 

Which simplifies to

 

 

Week 4 in Pre-Calculus 11 – Dividing Radicals

I chose this because its important to understand the rules to be able to complete the operations correctly.

Dividing monomials is pretty simple, as you simply divide as you think you would divide, so in this post I will focus on dividing binomials, which includes using a conjugate.

What is a conjugate? The conjugate of 1+2 is 1-2, so it’s simply just switching the sign in the middle of an operation.

First, we need the conjugate of the denominator to cancel it out, as we can see the conjugate of 3+\sqrt{10} is 3-\sqrt{10}

Next, we multiple the nominator and denominator by the conjugate to cancel out the denominator.

So:

Now we need to properly distribute by multiplying as usual.

is also the same as

This can still be further simplified by multiplying top and bottom by -1, which will reverse the signs.

After that, we still need to distribute in the nominator, giving us:

is also the same as which is the same as

 

 

 

Week 3 in Pre-Calculus 11 – Adding and Subtracting Radicals

I chose to explain this topic because to me it is more complicated than multiplying radicals, so I think there will be more educational value for me to explain this.

The operations for adding and subtracting radicals are done with mixed radicals, so being able to convert from entire to mixed (and sometimes vise-versa) is crucial.

To be able to combine (through adding/subtracting) radicals, the index and radicand have to be the same.

As said before, we have to convert to mixed, which is done the same as before.

In mixed radical form:

After converting to mixed, we can see that the radicals have the same index and radicand as some other radical, so now we actually add and subtract. To do this we just combine the coefficients with like terms (remember, like terms in radicals: same index & radicand)

So, 3+2 is 5, and -4-2 is -6, meaning the answer is

 

 

In the case of operations on unsimplified mixed radicals, the mixed radical still has to be in simplest mixed radical form.

While these radicals are mixed, they still can be further simplified.

63 becomes root 7 with a coefficient of 3, but we can’t just ignore the other coefficient. to combine them, simply just multiply the coefficients. \frac{1}{3} multiplied by \frac{3}{1} is 1.

Do the same thing to the second radical, and you will find out that they have the same radicand in simplest forms, meaning you can further simplify.

5\sqrt{45}

Week 2 in Pre-Calculus 11 – Writing Expression with Exponents as Entire Radicals

I chose this topic because while it covers new forms of radicals such as mixed and entire, it also reviews previous topics such as exponent laws.

To be able to solve the following examples, you need pre-existing knowledge on order of operations, exponent laws, and fractions.

  • Zero exponents: = 1
  • Negative exponents: = (reciprocal)
  • Product of exponents:
  • Quotient of exponents:
  • Power of a power:
  • Order of operations: Brackets, exponents, division/multiplication, addition/subtraction.

 

First, simplify out of the brackets. Looking at the exponents, use the power of a power rule to multiply. 3/2 * 2/1 = a^6/2 = a^3/1

Now look to completely simplify the first variable. Using the quotient of exponents rule, subtract. a^3 – a^3 = a^0 = 1

Simplify the second variable. b^3/1 = b^6/2, b^1/2 – b^6/2 = b^-5/2

Use negative exponents rule to simplify negative exponent to a positive. b^-5/2 = 1/b^5/2

Finally, recognize that fractional exponents show the radicand, index, and power for an entire radical. In this case, b^5/2 =

So, the answer is:

Week 1 in Pre-Calculus 11 – Radicals

This week I have learnt of a new term: radical. I chose to make a blog post about radicals because it’s the first new thing I have learnt in the year which will probably show up again.  A radical is an operation which includes a coefficient, index, root, and radicand.

For example,

In this radical, the 2 is the coefficient, which means multiply. The 3 is the index, which tells us what kind of root we’re looking for (in this case, cube). The 8 is the radicand, which is the number that you need to find the root of.

To solve the radical, you first find the cube root of 8 by thinking of how you can reach 8 through multiplying the same number by itself three times. Then, you look at the coefficient and multiply.

2*2*2=8

(2)2=4

It should be noted that when it appears as if there is no coefficient or index, the default value of the coefficient is 1, and the index’s is 2 (square).