Square Root Of 7: Value 2.6458, Methods And Proof

Where Does $\sqrt{7}$ Show Up, And Why Does It Matter?

The square root of 7 appears whenever a length squares to 7. One clean case is a right triangle with hypotenuse 4 and one leg 3, whose other leg is $\sqrt{16 - 9} = \sqrt{7}$.

It also hides around the equilateral triangle. The smallest rectangle that encloses an equilateral triangle of side 2 measures 2 by $\sqrt{3}$, and its diagonal is $\sqrt{4 + 3} = \sqrt{7}$ (Wikipedia, Square root of 7).

Triangles with a 120 degree angle produce it too. With sides 1 and 2 around a 120 degree angle, the law of cosine makes the third side $\sqrt{7}$, which Example 6 below works through.

Quick Reference Table

Seven is three steps past the perfect square 4 and two short of 9. The neighbouring roots below show where $\sqrt{7}$ fits.

For the rest of the range, see the square root 1 to 30 table.

What Is A Square Root?

A square root of $n$ is a number $r$ with $r^2 = n$. Both $\sqrt{7}$ and $-\sqrt{7}$ square to 7, so $x^2 = 7$ has two solutions.

The radical sign on its own means the principal (non-negative) root. So $\sqrt{7} \approx 2.6458$, and the negative root is written $-\sqrt{7} \approx -2.6458$.

In exponent form, $\sqrt{7} = 7^{\frac{1}{2}}$. The exponent $\frac{1}{2}$ means "square root", not "half of 7".

Is The Square Root Of 7 Rational Or Irrational?

The square root of 7 is irrational. No fraction of whole numbers equals it, so it is not a rational number, and its decimal runs on without repeating.

The quick reason: a whole number has a rational square root only when it is a perfect square. Seven lies strictly between the perfect squares 4 and 9, so $\sqrt{7}$ must be an irrational number.

Why Is The Square Root Of 7 Irrational?

The proof is a contradiction, and it works because 7 is one of the prime numbers.

1. Suppose $\sqrt{7} = \frac{p}{q}$ with whole numbers $p$ and $q$ that share no common factor.

2. Square both sides: $p^2 = 7q^2$, so 7 divides $p^2$.

3. A prime that divides $p^2$ must divide $p$, so write $p = 7k$.

4. Then $49k^2 = 7q^2$, which gives $q^2 = 7k^2$.

5. So 7 divides $q^2$, and therefore 7 divides $q$.

6. Now 7 divides both $p$ and $q$, which contradicts step 1.

The full write-up, line by line, is on the page that shows how to prove that root 7 is irrational.

How Do You Find The Square Root Of 7?

Five hand methods reach $2.6458$, each from a different direction. Prime factorization is the one that does not give a decimal, because 7 has no factor pairs to pull out.

Method 1: Long Division

Write 7 as $7.00,00,00,00,00$. At each step, double the root found so far, attach a trial digit $d$, and pick the largest $d$ whose product still fits.

At step 2 the digit cannot be 7, because $47 \times 7 = 329$ is more than 300. The remainder never reaches zero, so $\sqrt{7} \approx 2.64575$ and the digits continue.

Method 2: The Average Method

Pick a guess, divide 7 by it, and average the guess with the result. A guess that is too big produces a quotient that is too small, so the average lands in between.

$$x_{\text{new}} = \frac{1}{2}\left(x + \frac{7}{x}\right)$$

1. Start at $x = 3$: $\frac{7}{3} \approx 2.3333$, and $\frac{1}{2}(3 + 2.3333) = \frac{8}{3} \approx 2.6667$.

2. Use $\frac{8}{3}$: $\frac{1}{2}\left(\frac{8}{3} + \frac{21}{8}\right) = \frac{127}{48} \approx 2.645833$.

3. Use $\frac{127}{48}$: the next value is about $2.6457513123$, already correct to 8 decimal places.

Remember to halve the sum, because stopping at $2.33 + 3 = 5.33$ gives a number whose square is about 28.

This average method is the Babylonian method again, and it is also the Newton Raphson method applied to $x^2 - 7 = 0$.

Method 3: Squaring Estimates

Guess a decimal, square it, and tighten the range.

1. $2.6^2 = 6.76$ and $2.7^2 = 7.29$, so $\sqrt{7}$ is between 2.6 and 2.7.

2. $2.64^2 = 6.9696$ and $2.65^2 = 7.0225$, so it is between 2.64 and 2.65.

3. $2.645^2 = 6.996025$, still under 7, so the root is just above 2.645.

4. $2.6457^2 = 6.99972849$ and $2.6458^2 = 7.00025764$, so $\sqrt{7} \approx 2.6458$ to 4 places.

Method 4: A Binomial Shortcut

Write 7 as $\frac{63}{9}$, and notice that 63 is one less than 64. Then $\sqrt{7} = \frac{\sqrt{63}}{3} = \frac{8}{3}\sqrt{1 - \frac{1}{64}}$.

The binomial theorem for a small $x$ gives $\sqrt{1 - x} \approx 1 - \frac{x}{2} - \frac{x^2}{8}$. With $x = \frac{1}{64}$:

$$\sqrt{7} \approx \frac{8}{3}\left(1 - \frac{1}{128} - \frac{1}{32768}\right)$$

$$\approx 2.6457520$$

That is within $0.0000007$ of the true value, using only fractions.

Method 5: Prime Factorization

Seven is prime, so its factorization is just $7$. With no pair of equal factors, nothing comes out of the root, and $\sqrt{7}$ is already in simplest form.

The same idea simplifies larger roots that contain 7, such as $\sqrt{28} = 2\sqrt{7}$, which the guide to simplifying radical expressions covers.

How Do You Represent The Square Root Of 7 On A Number Line?

You can mark $\sqrt{7}$ exactly with a ruler and compass, even though its decimal never ends. Three constructions work, and each relies on the Pythagoras relationship.

Construction 1: The Semicircle Method

1. Mark $O$ at 0, $B$ at 7 and $C$ at 8 on a number line.

2. Find the midpoint $M$ of $OC$, which is at 4, and draw a semicircle on $OC$ with radius 4.

3. Draw a perpendicular to the line at $B$, meeting the semicircle at $D$.

4. In right triangle $MBD$, $MD = 4$ and $MB = 3$, so $BD = \sqrt{16 - 9} = \sqrt{7}$.

5. Set the compass to length $BD$, place the point at $O$, and mark the number line. That mark is $\sqrt{7}$.

Construction 2: A Right Triangle With Hypotenuse 4

1. Mark $A$ at 3 and draw a perpendicular to the line at $A$.

2. With the compass point at $O$ and radius 4, cut the perpendicular at $P$.

3. Then $AP = \sqrt{4^2 - 3^2} = \sqrt{7}$, since $OP$ is the hypotenuse of right triangle $OAP$.

4. Transfer $AP$ to the number line from $O$ with the compass.

Construction 3: The Spiral Method

Start with a right triangle whose legs are both 1, giving a hypotenuse of $\sqrt{2}$. Attach a new leg of length 1 at right angles to each hypotenuse, and the hypotenuses run $\sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}$.

After six triangles, swing the last hypotenuse down onto the number line with a compass. This growing staircase of triangles is called the spiral of Theodorus.

Can The Square Root Of 7 Be Written As A Fraction?

Not exactly, because it is irrational. Fractions can only approximate it, and the best ones come from its continued fraction $[2; 1, 1, 1, 4, \ldots]$.

The fractions $\frac{8}{3}$ and $\frac{127}{48}$ miss $p^2 = 7q^2$ by exactly 1, which is why the average method produced them. No fraction ever makes that last column zero.

Examples Of Square Root Of 7

Example 1

Simplify $3\sqrt{7} + 5\sqrt{7}$.

$$3\sqrt{7} + 5\sqrt{7} = 8\sqrt{7}$$

$$\approx 8 \times 2.6457513$$

$$\approx 21.1660$$

Example 2

Rationalise $\frac{14}{\sqrt{7}}$.

A tempting shortcut is to cancel the 7 inside the root against the 14: $\frac{14}{\sqrt{7}} = 2$.

Check with decimals: $14 \div 2.6458 \approx 5.29$, not 2. The 7 under a root is not the same as a 7 outside it.

Multiply top and bottom by $\sqrt{7}$ instead:

$$\frac{14}{\sqrt{7}} = \frac{14\sqrt{7}}{7}$$

$$= 2\sqrt{7}$$

$$\approx 5.2915$$

Example 3

Simplify $\sqrt{28}$, $\sqrt{63}$ and $\sqrt{175}$.

$$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \approx 5.2915$$

$$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7} \approx 7.9373$$

$$\sqrt{175} = \sqrt{25 \times 7} = 5\sqrt{7} \approx 13.2288$$

Each one pulls out a perfect square and leaves the same $\sqrt{7}$ behind.

Example 4

A right triangle has a hypotenuse of 4 cm and one leg of 3 cm. Find the other leg.

$$b = \sqrt{4^2 - 3^2}$$

$$= \sqrt{16 - 9}$$

$$= \sqrt{7} \approx 2.6458 \text{ cm}$$

Example 5

Solve $2x^2 - 14 = 0$.

$$2x^2 = 14$$

$$x^2 = 7$$

$$x = \sqrt{7} \text{ or } x = -\sqrt{7}$$

$$x \approx 2.6458 \text{ or } x \approx -2.6458$$

Example 6

Two sides of a triangle are 1 and 2, with a 120 degree angle between them. Find the third side.

By the law of cosines, with $\cos 120^\circ = -\frac{1}{2}$:

$$c^2 = 1^2 + 2^2 - 2 \times 1 \times 2 \times \cos 120^\circ$$

$$= 1 + 4 + 2$$

$$= 7$$

$$c = \sqrt{7} \approx 2.6458$$

Which Mistakes Cost Marks On The Square Root Of 7?

Mistake 1: Reading $7^{\frac{1}{2}}$ As Half Of 7

Where it slips in: Switching between radical form and exponent form.

Don't do this: Writing $7^{\frac{1}{2}} = 3.5$.

The correct way: The exponent $\frac{1}{2}$ means square root, so $7^{\frac{1}{2}} = \sqrt{7} \approx 2.6458$. A quick check catches it: $3.5^2 = 12.25$, which is not 7.

Mistake 2: Forgetting To Halve In The Average Method

Where it slips in: The averaging step, right after dividing 7 by the guess.

Don't do this: Adding $3 + 2.33 = 5.33$ and reporting that as the root.

The correct way: Divide the sum by 2 to get about 2.67, then square it as a check. The rusher who skips the halving also skips the check, and the check would show $5.33^2 \approx 28.4$ straight away.

Mistake 3: Rounding Before The Last Step

Where it slips in: Multiplying a rounded root, as in $7\sqrt{7}$.

Don't do this: Using $2.645$ and writing $7\sqrt{7} = 18.515$.

The correct way: Keep $\sqrt{7}$ exact until the end: $7\sqrt{7} \approx 18.5203$. A root rounded to three places loses accuracy every time it is multiplied, and students who keep the radical until the final line stop losing those marks.

Where Can You Get Help With Square Roots And Radicals?

Irrational roots, radical simplification and number-line constructions are where arithmetic turns into algebra. A teacher who works through each step with you keeps those ideas from blurring together.

You can work one-to-one with an algebra tutor, book a middle school math tutor for grades 6 to 8, or join math classes online for a steady weekly routine.

Conclusion

• The square root of 7 is about 2.6458 and stays as $\sqrt{7}$ in simplest form, since 7 is prime.

• It is irrational because 7 is not a perfect square; the contradiction proof uses the fact that 7 is prime.

• Long division, the average method, squaring estimates and the binomial shortcut all approach 2.64575131.

• A semicircle on 0 to 8, a right triangle with hypotenuse 4, or the spiral of Theodorus marks $\sqrt{7}$ exactly on a number line.

• Fractions such as $\frac{8}{3}$ and $\frac{127}{48}$ come close, but no fraction is exact.

Practice These Before Moving On

1. Find $\sqrt{7} \times \sqrt{3}$ to 4 decimal places.

2. Simplify $\sqrt{112}$.

3. Rationalise $\frac{21}{\sqrt{7}}$.

4. Expand $(\sqrt{7} + 2)(\sqrt{7} - 2)$.

5. Square $2.65$ and say whether $\sqrt{7}$ is above or below it.

Answers:

• Answer to Question 1: $\sqrt{21} \approx 4.5826$

• Answer to Question 2: $4\sqrt{7} \approx 10.5830$

• Answer to Question 3: $3\sqrt{7} \approx 7.9373$

• Answer to Question 4: $7 - 4 = 3$

• Answer to Question 5: $2.65^2 = 7.0225$, so $\sqrt{7}$ is just below 2.65

If Question 3 gave you 3, go back to Example 2 and redo the rationalising step. Want a teacher to check your working? Book a free demo class.

Read More

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