How do you solve sqrt (6x-5)+10=3? - Socratic √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
How do you factor #6x^2+x-1# - Socratic 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
Question #9adb8 - Socratic 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
Question #dec41 - Socratic 6x-y=11 The equation of a line in color (blue)"point-slope form" is color (red) (bar (ul (|color (white) (2 2)color (black) (y-y_1=m (x-x_1))color (white) (2 2
Question #6aa6a - Socratic (x+6) (x-6) >"factorise numerator denominator" • x^2+12x+36 "'split '12x into 6x + 6x" rArrx^2+6x+6x+36 "factor each 'pair' by taking out a "color (blue)"common