You can put this solution on YOUR website! 3x^2 - 2x - 3 = 0 ----(1)
Multiplying by 3
9x^2-6x-9 =0 ----(2) (zero mulitplied by anything is zero)
(3x)^2- 2X(3x)-9= 0
That is (3x)^2- 2X(3x)X(1)= 9
[The left hand side is like (a)^2 -2ab with a =(3x) and b = 1. we require b^2 to make it a perfect square: (a)^2 -2ab +b^2 = (a-b)^2 ]
Adding b^2 = 1^2 = 1 on both the sides
(3x)^2- 2X(3x)X (1) +1 = 9 +1
(3x-1)^2 = 10
Taking the sqrt on both the sides
(3x-1) = +or- sqrt(10)
3x = 1+or- sqrt(10)
x = (1/3)times[1+or- sqrt(10)]
Answer: x = (1/3)(1+sqrt10) and x= (1/3)(1-sqrt10)
Verification: x = (1/3)(1+sqrt10) in (1) implies
LHS = 3x^2 - 2x - 3
= 3X(1/9)(1+10+2sqrt10)-2X(1/3)X(1+sqrt10)-3
=(1/3)[11+2sqrt10-2(1+sqrt10)]-3
=(1/3)[11+2sqrt10-2-2sqrt10)]- 3
=(1/3)(11-2) - 3
=(1/3)X(9) - 3
=3-3 = 0 =RHS
Therefore x = (1/3)(1+sqrt10) is correct
Since surd roots occur in conjuate pairs in equations
there is no need for us to test the validity of the other root.
Note: Why did we multiply through out by 3 in the beginning?
Abs: To make the first term a perfect square
to play the role of a in the perfect square (a-b)^2
3x^2 - 2x - 3 = 0
Please help me solve this equation
By completing the square
3x² - 2x - 3 = 0
Get constant term on right by adding 3 to both sides
3x² - 2x = 3
Divide every term by the coefficient of x², namely 3
3 2 3
--- x² - --- x = ---
3 3 3
2
x² - --- x = 1
3
or written on one line
x² - (2/3)x = 1
Multiply coefficient of x, -2/3 by 1/2, getting -1/3
Square -1/3: (-1/3)2 = 1/9
Add 1/9 to both sides:
x² - (2/3)x + 1/9 = 1 + 1/9
The left side factors as
(x - 1/3)(x - 1/3) = 1 + 1/9
or
(x - 1/3)² = 1 + 1/9
The right side becomes 1 + 1/9 = 9/9 + 1/3 = 10/3
(x - 1/3)² = 10/9
Take the square root of both sides, remembering ±
on the right side
____
x - 1/3 = ±Ö10/9
We can take the square root of denominator 9 as 3
__
x - 1/3 = ±Ö10/3
Solving for x by adding 1/3 to both sides
__
x = 1/3 ± Ö10/3
Both fractions have same denominator, so combine
numerators over the common denominator
__
x = (1 ± Ö10)/3
Edwin
AnlytcPhil@aol.com
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