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√6x − 5+10 −10 = 3 −10 ⇒ √6x −5 = −7 square both sides (√6x − 5)2 = (− 7)2 ⇒ 6x − 5 = 49 add 5 to both sides. 6x−5 +5 = 49+ 5 ⇒ 6x = 54 divide both sides by 6 6x 6 = 54 6 ⇒ x = 9 As a check Substitute this value into the left side of the equation and if equal to the right side then it is the solution. This means that we can write the equation as 6x^2+x-1 = (2x+1) (3x-1) Here are a couple of methods (in no particular order): Method 1 Note that: (ax+1) (bx-1) = abx^2+ (b-a)x-1 Comparing with: 6x^2+x-1 we want to find a, b such that ab=6 and b-a = 1 The values a=2, b=3 work, so we find: 6x^2+x-1 = (2x+1) (3x-1) color (white) () Method 2 - Completing the square To avoid much arithmetic with fractions, multiply first by 24 = 6*2^2 then.

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We're dividing a single term, #6x^3#, into a polynomial, so we can take advantage of the following fact which allows us to effectively divide one term into other terms: Next we can factor out the area in parentheses How do you use the important points to sketch the graph of y = x2 − 6x + 1

Algebra quadratic equations and functions quadratic functions and their graphs

What is the numerical coefficient of the variable and the constant term of 6x + 5 What is the difference between a constant and a coefficient What is the coefficient of 2x In the term 5x, what is the coefficient?

This is equation [3] we found earlier. Explanation: The problem #int (2x)/ (x^2+6x+13)dx# can be solved By the method as shown below. Let #u=x^2+6x+13#, #du= (2x+6)dx# Further, in the numerator, we have only #2x=2x+6-6# the problem can be restated as Integrate the expression #int (2x+6-6)/ (x^2+6x+13)dx# By applying sum rule, #int (2x+6-6)/ (x^2+6x+13)dx=int (2x+6)/ (x^2+6x+13)dx-int (6)/ (x^2+6x+13)dx# Let #I_1=int (2x+6)/ (x^2+6x. They are additive inverses (i.e The same distance from 0, but on opposite sides), and the sum of two additive inverses is 0

We can combine the 15 and the 21 on the right side.

0, 1, 3 to solve this, we set the equation equal to 0 and factor it