Solving Equations with a Square Root

When we solve an equation by taking a square root, we have to consider both a positive and negative outcome.

For example consider -3 and 3

now 3² = 9 and -3²=9

Thus when we have an equation such that

x² = 9 to find x we need to take the square root of 9, however, then we have to say x= + or - 3 that is to say positive or negative 3 since it could be either and we don't know.

Multiplying and dividing square roots (radicals)

Square roots act just like other numbers when you multiply and divide them, consider the example

3sqrt(5) * 5sqrt(6) = 15sqrt(30)

Similarly

15sqrt(30) / 5sqrt(6) = 3sqrt(5)

Adding and Subtracting Radicals

When adding and subtracting radicals you treat the radicands as variables.

Example

3sqrt(5) + 4sqrt(5) = 7sqrt(5)

However, we cannot add together a radicand that is different as in

3sqrt(5) + 4sqrt(2)

The square root of a fraction

By definition we cannot derive the square root of a fraction. Thus we must find a way to get a fraction out of the radical.

Consider the square root of 1/2

sqrt(1/2)

how can we get the fraction out? We have to multiply and make the denominator a perfect square. But what we multiply to the bottom we must also multiply to the top

sqrt((1/2)*(2/2))

=

1/2*sqrt(2)

Now we have got the fraction out of the radical and created a square root we can rationalize.

Another example

sqrt(3/5)

To get the denominator out we multiply by (5/5) (which is equal to 1)

sqrt((3/5)*(5/5))

=

(1/5)*sqrt(3*5)

=

(1/5)*sqrt(15)

Square roots are radical

It is true, taking the square root of an expression can also be called "The radical"

We will write square root as sqrt on this website, thus

sqrt(4) = 2

and

sqrt(9x²y¹⁰) ?

To solve this, it is good to factor under the radical ( or factor the square root)

So we get

sqrt(9x²y¹⁰)

=

(3xy⁵)

Prove that the square root of 2 is irrational

Prove that the square root of 2 is irrational.

We will use the well ordering principle for this proof and try to reach a contradiction.

Let A be a set in the natural numbers so that there exists b and a in the natural numbers so that sqrt(2)=(a)/(b)

thus 2 =(a^2)/(b^2)

and (a^2)= 2(b^2)

From the formula above we see that a must be an even number because it is equal to 2(b^2). If we substitute 2k for a to represent this we get:

(2k^2) = 2 (b^2)

(b^2) = 4(k^2) / 2 = 2(k^2)

so
b = 2k

And so b also must be even, however if both a and b are smallest elements in the set A, then one must be even and one must be odd so we have a contradiction and the square root of 2 cannot be expressed as the ratio of two numbers, thus it is irrational.