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- 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 use the quadratic formula to solve 6x ^ { 2} - 24x - 30= 0 . . .
Explanation: After play with some of the pairs to combine #6, 24 and 30# #6x^2 - 24x - 30 = 0# can be factored to:
- 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
- Question #c2ea6 - Socratic
1 Answer Sunayan S Feb 14, 2018 Future reference:- (see explanation first) From, # (2)# and # (3)#,
- What are the real zeros of this function #f (x)=3x^3+6x^2 . . . - Socratic
f (x) has zeros +-sqrt (5) and -2 Given: f (x) = 3x^3+6x^2-15x-30 Note that all of the terms are divisible by 3 In addition, the ratio of the first and second terms is the same as that of the third and fourth terms
- Question #919b3 - Socratic
36x^2-25=(6x+5)(6x-5) Using the difference of squares formula a^2-b^2 = (a+b)(a-b), we have 36x^2-25 = 6^2x^2-5^2 =(6x)^2-5^2 =(6x+5)(6x-5)
- 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
- How do you solve this system of equations: 3y + 6x = - 24 . . . - Socratic
Using [1] 3y+6x=-24 3y=-24-6x color (blue) (y=-8-2xcolor (white) (888) [3] Substituting [3] in [2] z=-2 (-8-2x)-16 z=16+4x-16 z=4x Now find y in terms of x This is equation [3] we found earlier
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