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This page provides a list of educational videos related to Squaring Numbers. You can also use this page to find sample questions, apps, worksheets, lessons , infographics and presentations related to Squaring Numbers.
Square Roots and Real Numbers
By Khan Academy
Khan Academy presents Square Roots and Real Numbers an educational video resource on math.
Square roots and real numbers
By MathPlanetVideos
Determine whether these numbers are rational or irrational
Quadratic Word Problems | MathHelp.com
By MathHelp.com
A number is 56 less than its square. Find the number. To solve this problem, let’s translate the first sentence into an equation. A number, that’s x, is, =, 56 less than it’s square, that’s x squared – 56. Remember that “less than” switches the order around. In other words, “56 less than its square” is not 56 minus x squared, it’s x squared minus 56. Next, since we have an x squared term in our equation, we set it equal to 0 by subtracting x from both sides, and we have 0 = x squared – x – 56. Next, we factor the right side as the product of two binomials. In the first position of each binomial, we have the factors of x squared, x and x. In the second position of each binomial, we’re looking for the factors of -56 that add to -1, which are -8 and positive 7. So we have 0 = x - 8 times x + 7, which means that either 0 = x – 8 or 0 = x + 7. Finally, in the first equation, we add 8 to both sides, to get 8 = x. And in the second equation, we subtract 7 from both sides, to get -7 = x. So 8 = x or -7 = x. It’s important to understand that both of these answers work. Plugging an 8 back into the original problem, we have 8 is 56 less than 8 squared, or 8 = 8 squared – 56, which simplifies to 8 = 64 – 56, or 8 = 8, which is a true statement. And plugging a -7 back into the original problem, we have -7 is 56 less than -7 squared, or -7 = -7 squared – 56, which simplifies to -7 = 49 – 56, or -7 = -7, which is also a true statement.
Approximating square roots
By Khan Academy
Learn how to find the approximate values of square roots. The examples used in this video are √32, √55, and √123. The technique used is to compare the squares of whole numbers to the number we're taking the square root of.
Imaginary Numbers | MathHelp.com
By MathHelp.com
This lesson covers estimating quotients. Students learn to estimate a product or quotient by first rounding each number to one non-zero digit. For example, to estimate 8,291 x 27, first round 8,291 down to 8,000, and round 27 up to 30, then multiply 8,000 x 30 to get 240,000.
Parallel Lines | MathHelp.com
By MathHelp.com
This lesson covers imaginary numbers. Students learn that the imaginary number "i" is equal to the square root of -1, which means that i^2 is equal to (the square root of -1) squared, which equals -1. Students also learn to simplify imaginary numbers. For example, to simplify the square root of -81, think of it as the square root of -1 times the square root of 81, which simplifies to i times 9, or 9i. To simplify 11/8i, the first step is to get rid of the "i" in the denominator by multiplying both the numerator and the denominator of the fraction by i, to get 11i/8i^2, and remember that i^2 = -1, so we have 11i/8(-1), or 11i/-8, or -11i/8.
Solving Logarithmic Equations | MathHelp.com
By MathHelp.com
Here we’re asked to evaluate each of the following logarithms. In part a, we have log base 7 of 49. To evaluate this logarithm, we set it equal to x. In other words, log base 7 of 49 = what? Notice that we now have an equation written in logarithmic form, so let’s see if we can solve the equation by converting it to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 7…to the x…= 49. Next, we solve for x. Notice that 7 and 49 have a like base of 7, so we rewrite 49 as 7 squared, and we have 7 to the x = 7 squared, so x must equal 2. In part b, we have log base 3 of 1/27. Again, to evaluate this logarithm, we set it equal to x, and convert the logarithmic equation to exponential form. Remember that the base of the log represents the base of the power, the right side of the equation represents the exponent, and the number inside the log represents the result, so we have 3…to the x…= 1/27. Next, we solve for x. Notice that 3 and 1/27 have a like base of 3, so we rewrite 1/27 as 1 over 3 cubed, and we have 3 to the x = 1 over 3 cubed. Next, 1 over 3 cubed is the same thing as 3 to the negative 3, so we have 3 to the x = 3 to the negative 3, which means that x must equal -3. Therefore, log base 3 of 1/27 = -3. So remember the following rule. To evaluate a logarithm, set it equal to x, convert to exponential form, and solve the equation using like bases.