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Jesse Womble
Jesse Womble

How To Solve Square Roots Problems In Common Core Algebra 1 [Extra Quality]



How to Solve Square Roots Problems in Common Core Algebra 1




If you are taking Common Core Algebra 1, you may encounter some problems involving square roots. Square roots are numbers that, when multiplied by themselves, give another number. For example, the square root of 25 is 5, because 5 times 5 equals 25. In this article, we will show you how to solve some common types of square roots problems in Common Core Algebra 1.




How to Solve Square Roots Problems in Common Core Algebra 1



Problem 1: Simplifying Square Roots




One type of problem you may face is simplifying square roots. This means finding the simplest way to write a square root expression. For example, how can we simplify the square root of 50? To do this, we need to find the largest perfect square that divides 50. A perfect square is a number that has a whole number as its square root, such as 4, 9, 16, and so on. The largest perfect square that divides 50 is 25, because 25 times 2 equals 50. So, we can write the square root of 50 as the product of the square root of 25 and the square root of 2. Then, we can simplify the square root of 25 to 5, since we know that 5 times 5 equals 25. Therefore, the square root of 50 is equal to 5 times the square root of 2, or 5√2.


Here is another example: how can we simplify the square root of 18? The largest perfect square that divides 18 is 9, because 9 times 2 equals 18. So, we can write the square root of 18 as the product of the square root of 9 and the square root of 2. Then, we can simplify the square root of 9 to 3, since we know that 3 times 3 equals 9. Therefore, the square root of 18 is equal to 3 times the square root of 2, or 3√2.


Problem 2: Solving Equations with Square Roots




Another type of problem you may face is solving equations with square roots. This means finding the value of a variable that makes an equation true. For example, how can we solve this equation: x + √x = -3? To do this, we need to isolate x on one side of the equation. We can start by subtracting x from both sides: √x = -3 - x. Then, we can square both sides to get rid of the square root: x = (-3 - x)^2. Next, we can expand the right side using the formula (a - b)^2 = a^2 - 2ab + b^2: x = (-3)^2 - 2(-3)(-x) + (-x)^2. Then, we can simplify the right side by performing the operations: x = 9 +6x + x^2. Next, we can move all the terms to one side by subtracting x from both sides: 0 = x^2 +5x -9. Finally, we can factor the left side using the method of trial and error: 0 = (x +9)(x -4). Now, we can use the zero product property to find the values of x that make each factor equal to zero: x +9 =0 or x -4 =0. Solving for x, we get: x =-9 or x =4.


However, we are not done yet. We need to check if these solutions are valid by plugging them back into the original equation. If we plug in x =-9, we get: -9 + √(-9) = -3. This is not true, because the square root of a negative number is not a real number. So, x =-9 is an extraneous solution that we need to reject. If we plug in x =4, we get 04f6b60f66


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