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3 years ago

Solve the following equation algebraically:

Mathematics

1 answer:

Gre4nikov [31]3 years ago

8 0

Answer:

$\boxed{\sf \ \ \ answer \ b. \ \ +5 \ \ \ }$

Step-by-step explanation:

Hello

let's assume that we are looking for x >=0

$4x^2=100 \dfrac{4x^2}{4}=x^2=\dfrac{100}{4}=25\\\sqrt{x^2}=25\\ |x| = 5 x = 5 \ or \ x=-5$

Let's assume that we are looking for x>=0

so the correct answer is 5

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2. Solve the equation by completing the square. Show your work.

scoundrel [369]

Answer:

x = 5 or 25

Step-by-step explanation:

I think the method I am about to explain is slightly quicker and easier than the method in your question. This works for any 'complete the square' question.

We begin with x² - 30x = - 125.

First, we are going to factorise the left-hand side of the equation by dividing the 'b' value (-30) by 2 (you'll see why this works in a minute):

(x - 15)²

We want these brackets to multiply out to give x² - 30x, so that they equal the left-hand side of the equation. Unfortunately, if we multiply them out, we get:

(x - 15)(x - 15) =

x² - 30x + 225

There is an unwanted term (the + 225, from 15²)! We only want x² - 30x, so we have to remove this term by subtracting it from the left side of the equation. To do this, let's set up the original equation again:

(x² - 30x + 225) - 225 = - 125

Note: The reason why we don't have to subtract it from both sides is because the original equation is x² - 30x = - 125, and so we must make sure the left hand side is still equal to x² - 30x.

So now we know that (x - 15)² multiplies out to give x² - 30x +225, we can write this as (x - 15)² in our equation:

(x² - 30x + 225) - 225 = - 125

is the same as:

(x - 15)² - 225 = - 125

Now add 225 to both sides of the equation:

(x - 15)² = - 125 + 225 = 100

(x -15)² = 100

The next step is to square root both sides. Be careful here, and remember that √100 can either be 10 or -10, as (-10)² = 100. To indicate both results, write ±10 ("plus or minus 10").

√(x - 15)² = √100

x - 15 = ±10

Because, there are two possible values for the right-hand side of the equation, we need to separate our equation into two equations:

1. x - 15 = 10

and

2. x - 15 = -10

Now we solve these two simple linear equations for x:

1. x = 10 + 15 <- add 15 to both sides

x = 25 This is our first solution.

2. x = -10 + 15 <- add 15 to both sides again

x = 5 This is our other solution.

So our two solutions are x = 5 and x = 25!

I have attached the quick version of the working out for this question - that is what you would be expected to write down in a test.

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4 years ago