Solve without using the quadratic formula?

Question

A) 2(x-3/4)^2/3 = 1/6

B) x^2+4x+4=9

C) 3(4x2-4x+1)=12

Answer 1

A) (x-3/4)^2/3=1/12
(x-3/4)^2=1/1728 (cube both sides)
x-3/4=+/-1/√1728
x=3/4 + 1/√1728
or x= 3/4 - 1/√1728
B) x^2 +4x -5=0
x^2+5x-x-5=0
x(x+5)-1(x+5)=0
x= -5 or x=1
C) 4x^2-4x+1=4
4x^2-4x-3=0
4x^2-6+2x-3=0
2x(2x-3)+1(2x-3)=0
(2x-3)(2x+1)=0
x=3/2 or x= -1/2

Answer 2

THE first equation u should make it power 3 so u will get
8 (x-3/4)^2=1/216
by solving this u will get that x^2=15/16
the second one x^2+4x=5 take x as a common factor
so x (x+4)=5 giving x(x+4)-5=0 so x=0 or x+4-5=0 x-1=0 so x=zero or one
the third one has the same idea as the sec. just enter the 3 in the brackets and take 12x as common and u will do it

Answer 3

B) x^2 + 4x + 4 = 9

First: set the equation to "0" - subtract 9 from both sides (when you move a term to the opposite side, always use the opposite sign)....

x^2 + 4x + 4 - 9 = 9 - 9
x^2 + 4x - 5 = 0

Sec: factor > multiply the 1st & 3rd coefficient to get "-5." Find two numbers that give you "-5" when multiplied & "4" (2nd/middle coefficient) when added/subtracted. The numbers are (5 & -1).

*Rewrite the expression with the new middle coefficients....

x^2 + 5x - x - 5 = 0

*With 4 terms - group "like" terms & factor both sets of parenthesis....

(x^2 - x) + (5x - 5) = 0
x(x-1) + 5(x-1) = 0
(x-1)(x+5) = 0

Third: solve the x-variables > set both parenthesis to "0"....

a. x - 1 = 0
ISolate "x" on one side - add 1 with both sides...

x - 1 + 1 = 0 + 1
x = 1

b. x + 5 = 0
*Subtract 5 from both sides...

x + 5 - 5 = 0 - 5
x = - 5

Solutions: 1 & -5

Answer 4

a)
(x - 3/4)^2/3 = 1/12
(x - 3/4)^2 = 1/(12)^3
x - 3/4 = 1/(24*sqrt 3)
x - 3/4 = (sqrt 3)/72
x = 3/4 + (sqrt 3)/72
x = 54/72 + (sqrt 3)/72
x = (54 + sqrt 3)/72

b)
x^2 + 4x + 4 = 9
x^2 + 4x - 5 = 0
(x + 5)(x + 1) = 0

x = -1 and -5

c)
3(4x^2 - 4x + 1) = 12
4x^2 - 4x + 1 = 4
4x^2 - 4x - 3 = 0
(2x - 3)(2x + 1) = 0

x = 3/2 and -1/2

Answer 5

Express each equation in the form:

something times x squared +/- something times x +/- something = 0

Call the three somethings a, b and c. Then look for two numbers which multiplied together give you a times c, and added together give you b.

Write the equation as (x - first number)(x - second number) = 0

Now you can say that (x - first number) = 0 or (x - second number) = 0.

Solved.