What is Square Root of 7?
The square root of a number is a value that, when multiplied by itself, gives us the original number. Hence, finding the square root is the converse of finding the square of a number.
The square root of \( 7 \) is written as \( \sqrt{7} \), with the radical sign ‘ \( \sqrt{} \) ‘and the radicand being \( 7 \). The square root of \( 7 \) has a value that is nearly equal to \( 2.64575 \), and this value is non terminating and non-repeating, showing us that it is an irrational number.
How did we obtain the value of \( \sqrt{7} \) as \( 2.64575\ldots? \) We use the long division method to find this value.
Deriving the Square Root of 7
We can apply the following steps to determine the square root of \( 7 \) using the long division method.
Step 1: Rewrite the number as shown below.
\( \overline{7}.\overline{00} ~\overline{00}~\overline{00} \)
Step 2: Take a number whose square is less than or equal to \( 7 \).
\( 2^2 =4\) , which is less than \( 7 \). So we will take \( 2 \).
Step 3: Write the number \( 2 \) as the divisor and \( 7 \) as the dividend. Now divide \( 7 \) by \( 2 \).
Here, Quotient \(= 2 \) and Remainder \( =3 \).
Step 4: Bring down 00 and write it after \( 3 \), so the new dividend is \( 300 \) and add the quotient \( 2 \) to the divisor, that is, \( 2+2=4 \).
Step 5: Add a digit right next to \( 4 \) to get a new divisor such that the product of a number with the new divisor is less than or equal to \( 300 \).
\( 46 \times 6=276 \) , which is less than \( 300 \).
Subtract the product \( 276 \) from \( 300 \) to get the remainder.
Here, Quotient \( =2.6 \) and Remainder \( =24 \)
Step 6: Repeat the previous two steps to obtain the quotient up to three decimal places.
Therefore, the value of the square root of \( 7 \), that is, \( \sqrt{7} \) is approximately \( 2.645 \).
Positive and Negative Square Root of a Number
Each number has two square roots, one is positive and other is negative. Let us understand this with example, multiply \( -\sqrt{7} \) by itself.
\( (-\sqrt{7})\times (-\sqrt{7})=\sqrt{7^2}=7 \) [Product of two negatives is a positive]
Also, multiply \( \sqrt{7} \) by itself
\( \sqrt{7}\times \sqrt{7}=\sqrt{7^2}=7 \)
From the above the square of \( -\sqrt{7} \) is \( 7 \) and square of \( \sqrt{7} \) is also \( 7 \), so we can say that square roots of \( 7 \) are \( \sqrt{7} \) and \( -\sqrt{7} \) both.
In general,
- \( \sqrt{y} \) represents the positive square root of \( y \).
- \( -\sqrt{y} \) represents the negative square root of \( y \).
- \( \pm \sqrt{y} \) represents both square roots of \( y \).
Perfect Squares and Imperfect Squares
The square root of a perfect square number is always an integer. In contrast, the value of the square root of imperfect squares is a non-integer, that is, it contains decimals or fractions.
For example, \( 2 \) is the square root of \( 4 \), which is a perfect square, and \( 1.414 \) is the square root of \( 2 \), which is an imperfect square.
Rapid Recall
Note: This table lists the approximate values of the square roots of some numbers that can be memorized to determine the square roots of higher imperfect square numbers.
Solved Square Root of 7 Examples
Example 1: Simplify and evaluate the expression:
\( \sqrt{14}-\sqrt{108-101} \)
Solution:
\( \sqrt{14}-\sqrt{108-101} \)
\( =\sqrt{14}-\sqrt{7} \)
\( =\sqrt{2\times 7}-\sqrt{7} \)
\( =\sqrt{2}\times \sqrt{7}-\sqrt{7} \)
\( =\sqrt{7}(\sqrt{2}-1) \)
\( =2.646(1.414-1) \) [Substitute \( 2.646 \) for \( \sqrt{7} \) and \( 1.414 \) for \( \sqrt{2} \) ]
\( =1.095444 \) [Solve]
Hence the simplified form of \( \sqrt{14}-\sqrt{108-101} \) is \( \sqrt{7}(\sqrt{2}-1) \) which is equal to \( 1.095444 \).
Example 2: Solve the equation given below:
\( \sqrt{583-(222+360)}~b-\sqrt{749-742}=0 \)
Solution:
\( \sqrt{583-(222+360)}~b-\sqrt{749-742}=0 \) Write the equation
\( \sqrt{583-582}~b-\sqrt{7}=0 \) Solve \( 222 + 360 \) and \( 749 – 742 \)
\( \sqrt{1}~b-\sqrt{7}=0 \) Solve \( 583 – 582 \)
\( b-\sqrt{7}=0 \) Substitute \( 1 \) for \( \sqrt{1} \)
\( b=\pm \sqrt{7} \) Solve for b
The solutions are \( b= \sqrt{7} \) and \( b= -\sqrt{7} \)
Example 3: Can you help Ava to find the value of hypotenuse of right angle triangle whose other two sides are \( \sqrt{5} \) in. and \( \sqrt{2} \) in.
Solution:
Assume \( \sqrt{5} \) as perpendicular length and \( \sqrt{2} \) as base length
Use pythagorean theorem to find the hypotenuse.
\( H^2 = P^2 + B^2 \)
\( H^2=(\sqrt{5})^2+(\sqrt{2})^2 \) Substitute \( \sqrt{5} \) for P and \( \sqrt{2} \) for B
\( H^2=5+2 \) Find the square of \( \sqrt{5} \) and \( \sqrt{2} \)
\( H^2=7 \) Add
\( \sqrt{H^2}=\sqrt{7} \) Apply square root on both sides
\( H=\sqrt{7} \) Simplify
So, the length hypotenuse is \( \sqrt{7} \) inches or about \( 2.646 \) inches.
The square root of an imperfect square can be calculated most precisely using the long division approach.
‘±’ stands for two results, one of which is positive and the other is negative with the same absolute value.
Square root of \( 7 \) in radical form is written as \( \sqrt{7} \).
No, the square root of \( 7 \) are \( \sqrt{7} \) and \( -\sqrt{7} \),which are irrational numbers.