.The square of a real number c is 11 less than the square of 5.

In an earlier chapter we learned that

32=3⋅3=9

We said that 9 was the square of 3. The square of -3 is 9 as well

(−3)2=(−3)⋅(−3)=9

3 and -3 are said to be the square roots of 9.

All positive real numbers has two square roots, one positive square root and one negative square root. The positive square root is sometimes referred to as the principal square root. The reason that we have two square roots is exemplified above. The product of two numbers is positive if both numbers have the same sign as is the case with squares and square roots

a2=a⋅a=(−a)⋅(−a)

A square root is written with a radical symbol √ and the number or expression inside the radical symbol, below denoted a, is called the radicand.

a−−√

To indicate that we want both the positive and the negative square root of a radicand we put the symbol ± (read as plus minus) in front of the root.

±9–√=±3

Zero has one square root which is 0.

0–√=0

Negative numbers don't have real square roots since a square is either positive or 0.

If the square root of an integer is another integer then the square is called a perfect square. For example 25 is a perfect square since

±25−−√=±5

If the radicand is not a perfect square i.e. the square root is not a whole number than you have to approximate the square root

±3–√=±1.73205...≈±1.7

The square roots of numbers that are not a perfect square are members of the irrational numbers. This means that they can't be written as the quotient of two integers. The decimal form of an irrational number will neither terminate nor repeat. The irrational numbers together with the rational numbers constitutes the real numbers.

Example

irrationalnumber⇒19−−√≈4.35889...

rationalnumber⇒0.5=12

step-by-step explanation: h: (- 3, 6)

m: (1, - 2)

o: (7, 6)

r: (- 9, - 2)

midpoint of hm: (- 1, 2)

midpoint of or: (- 1, 2)

therefore, b is the midpoint of both diagonals bisect each other !

to prove the diagonals are perpendicular, show that the product of the slopes = - 1

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .