This week in math 11, I learnt how to solve radical equations using factoring.

An equation in which a variable is in the radicand of a radical expression is called a radical equation.  To solve a radical equation: Isolate the radical expression involving the variable. If more than one radical expression involves the variable, then isolate one of them. let’s get into it!

As usual, when solving these equations, just like we learnt in grade 10. what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Step one is to isolate the radical on onside of the equation, in this case we move positive 4 over to the other side of the equation making the 4 negative and it nows reads = x – 4. our next step is to raise both sides of the equation to the power of the index. Since the index of a square root is 2, we square both sides. by doing this we now have a new equation with not radical. Solve the new equation, Remembering that (a)² = a. On the other side of the equal sign, since we have (x – 4)² we double the brakes and FOIL the equation. This will leave us with a trinomial on one side and a binomial on the other. Next we need to combine the binomial into the trinomial and have 0 on the other side. since we don’t want a negative x² we will subtract x and add 2 into the trinomial.

our send part is too factor our equation then solve for x, in this example I got two possibilities.

our last part is to verify our answer. 1st I tried x = 3, and by inputing 3 into x’s spot I found that x is equal to 5 and it does not match my answers, so the only possibility is x + 6.

hope this helps  🙂

 

 

This week in math 11, I relearned how to correctly factor trinomials. Learning how to factor expressions is important because it helps you answer your solution making it simpler to read and verify it so you know its correct.

A Trinomial is a polynomial that contains three terms. The first term is an x² term, the second is an x term, and the third is a constant (a single number).

ab+a+b     or      x²+x+12

To factor a trinomial, you need to find two numbers that can multiply the constant while also adding/subtracting from the coefficient of the variable with no apparent exponent. Once you have those two numbers, split them using brackets. This will also include the variable from the original expression, resulting in the same variable but different numbers in both brackets. To determine the signs that separate the numbers from the variable in each bracket, consider that when multiplying the numbers together, the sign should match the one on the constant, and when adding or subtracting the numbers, the sign and number of the coefficient with no visible exponent should match.

EXAMPLE 1)

 

In the example above, we have x squared, minus 7x, plus 12. There are three combinations can multiply to 12 including 12 x 1, 3 x 4, and 2 x 6. We can limit our options by deciding which of these can add or subtract to make the negative coefficient of 7. 3 plus 4 equals to 7 so that is the combination we are going to use. but we have to do the opposite integer to get negative 7. We are then going to create two brackets containing the variable, in this example x, and put one in each bracket and separate the combination of numbers we just decided on, – 3 and – 4, and put them in the same brackets as the variables. To determine the sign that will separate the constants from the variables, you want to think about the signs attached to the coefficient and constant in the original expression. Since the constant is 12 and the coefficient of x is -7. The signs both have to be negative so they can add up to -7 and multiply to become a positive to equal +12. To check you factored correctly, you can foil, expand it, or box method, and compare to the original expression, which should result in being the same.

EXAMPLE 2)

This example follows the exact same steps as example 1. This example just has exponents on two out of the three terms. When doing this question you want your brackets to have exponents that match the middle value.

EXAMPLE 3)

 

This example follows the same steps and example 1 and 2. But in this example there is a coefficient attached to the first term. To get ride of it for the time being you have to multiple the 3 into the third term to make it go away giving us 18. Then you continue with your normal steps finding number that both can be multiplied to 18 and added to 7. once you have your correct brackets there is one more step. you then divide your final numbers by the 3 from earlier. as you can see above the -9 can be divided by three giving us -3, but the two can’t be divided to get a whole number. so the three gets placed in-front of the t³. Giving us our final answer.

hope this helps 🙂