How To Solve Square Root Problems With Pictures

Try squaring a few more numbers on your own to test this concept out. Remember, squaring a number is just multiplying it by itself. You can even do this for negative numbers. If you do, the answer will always be positive. For example, (-8)2 = -8 × -8 = 64.

As another example, let’s find the square root of 25 (√(25)). This means we want to find the number that squares to make 25. Since 52 = 5 × 5 = 25, we can say that √(25) = 5. You can also think of this as “undoing” a square. For example, if we want to find √(64), the square root of 64, let’s start by thinking of 64 as 82. Since a square root symbol basically “cancels out” a square, we can say that √(64) = √(82) = 8.

On the other hand, numbers that don’t give whole numbers when you take their square roots are called imperfect squares. When you take one of these numbers’ square roots, you usually get a decimal or fraction. Sometimes, the decimals involved can be quite messy. For instance, √(13) = 3. 605551275464. . .

12 = 1 × 1 = 1 22 = 2 × 2 = 4 32 = 3 × 3 = 9 42 = 4 × 4 = 16 52 = 5 × 5 = 25 62 = 6 × 6 = 36 72 = 7 × 7 = 49 82 = 8 × 8 = 64 92 = 9 × 9 = 81 102 = 10 × 10 = 100 112 = 11 × 11 = 121 122 = 12 × 12 = 144

Let’s say that we want to find the square root of 900. At first glance, this looks very difficult! However, it’s not hard if we separate 900 into its factors. Factors are the numbers that can multiply together to make another number. For instance, since you can make 6 by multiplying 1 × 6 and 2 × 3, the factors of 6 are 1, 2, 3, and 6. Instead of working with the number 900, which is somewhat awkward, let’s instead write 900 as 9 × 100. Now, since 9, which is a perfect square, is separated from 100, we can take its square root on its own. √(9 × 100) = √(9) × √(100) = 3 × √(100). In other words, √(900) = 3√(100). We can even simplify this two steps further by dividing 100 into the factors 25 and 4. √(100) = √(25 × 4) = √(25) × √(4) = 5 × 2 = 10. So, we can say that √(900) = 3(10) = 30.

Note that although imaginary numbers can’t be represented with ordinary digits, they can still be treated like ordinary numbers in many ways. For instance, the square roots of negative numbers can be squared to give those negative numbers, just like any other square root. For example, i2 = -1

Start by writing out your square root problem in the same from as a long division problem. For example, let’s say that we want to find the square root of 6. 45, which is definitely not a convenient perfect square. First, we’d write an ordinary radical symbol (√), then we’d write our number underneath it. Next, we’d make a line above our number so that it’s in a little “box” — just like in long division. When we’re done, we should have a long-tailed “√” symbol with 6. 45 written under it. We’ll be writing numbers above our problem, so be sure to leave space.

In our example, we would divide 6. 45 into pairs like this: 6-. 45-00. Note that there is a “leftover” digit on the left — this is OK.

In our example, the first group in 6-. 45-00 is 6. The biggest number that is less than or equal to 6 when squared is 2 — 22 = 4. Write a “2” above the 6 under the radical.

In our example, we would start by taking the double of 2, the first digit of our answer. 2 × 2 = 4. Next, we would subtract 4 from 6 (our first “group”), getting 2 as our answer. Next, we would drop down the next group (45) to get 245. Finally, we would write 4 once more to the left, leaving a small space to add onto the end, like this: 4_.

In our example, we want to find the number to fill in the blank in 4_ × _ that makes the answer as large as possible but still less than or equal to 245. In this case, the answer is 5. 45 × 5 = 225, while 46 × 6 = 276.

Continuing from our example, we would subtract 225 from 245 to get 20. Next, we would drop down the next pair of digits, 00, to make 2000. Doubling the numbers above the radical sign, we get 25 × 2 = 50. Solving for the blank in 50_ × _ =/< 2,000, we get 3. At this point, we have “253” above the radical sign — repeating this process once again, we get a 9 as our next digit.

In our example, the number under the radical sign is 6. 45, so we would simply slide the point up and place it between the 2 and 5 digits of our answer, giving us 2. 539.

For example, let’s say we need to find the square root of 40. Since we’ve memorized our perfect squares, we can say that 40 is in between 62 and 72, or 36 and 49. Since 40 is greater than 62, its square root will be greater than 6, and since it is less than 72, its square root will be less than 7. 40 is a little closer to 36 than it is to 49, so the answer will probably be a little closer to 6. In the next few steps, we’ll narrow our answer down.

In our example problem, a reasonable estimate for the square root of 40 might be 6. 4, since we know from above that the answer is probably a little closer to 6 than it is to 7.

Multiply 6. 4 by itself to get 6. 4 × 6. 4 = 40. 96, which is slightly higher than original number. Next, since we over-shot our answer, we’ll multiply the number one tenth less than our estimate above by itself and to get 6. 3 × 6. 3 = 39. 69. This is slightly lower than our original number. This means that the square root of 40 is somewhere between 6. 3 and 6. 4. Additionally, since 39. 69 is closer to 40 than 40. 96, you know the square root will be closer to 6. 3 than 6. 4.

In our example, let’s pick 6. 33 for our two-decimal point estimate. Multiply 6. 33 by itself to get 6. 33 × 6. 33 = 40. 0689. Since this is slightly above our original number, we’ll try a slightly lower number, like 6. 32. 6. 32 × 6. 32 = 39. 9424. This is slightly below our original number, so we know that the exact square root is between 6. 33 and 6. 32. If we wanted to continue, we would keep using this same approach to get an answer that’s continually more and more accurate.