solve quadratic equation by completing the square 9x²-3x-2=0
Answers
Solution by completing the square for:
9x2−3x−2=0
a≠1 so divide through by 9
9x2/9−3x/9−2/9=0/9
which gives us
x2−1/3x−2/9=0
Keep x terms on the left and move
the constant to the right side
by adding it on both sides
x2−1/3x=2/9
Take half of the x term and square it
[−1/3⋅1/2]2=1/36
then add the result to both sides
x2−1/3x+1/36=2/9+1/36
Rewrite the perfect square on the left
(x−1/6)2=2/9+1/36
and combine terms on the right
(x−1/6)^2=1/4
Take the square root of both sides
x−1/6=√±14
Simplify the Radical term :
x−1/6=±1/2
Isolate the x on the left side and
solve for x
x=1/6±1/2
therefore
x=1/6+1/2 or x=1/6−1/2
which becomes
x=0.666667 or x=−0.333333
Answer:
x=2/3 or -1/3
Step-by-step explanation:
9x²-3x-2=0
Step I : Taking -2 to right side
9x²-3x=2
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STEP II
multiplying both sides by 4a=4*9=36
36*9x²-108x=72
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step III
adding both sides b²=(-3²)=9
36*9x²-108x+9=72+9
(18x)²-2*18x*3+3³=81
(18x-3)²=9²
Taking square root
18x-3=±9
6x-1=±3
6x=1±3
x=(1±3)/6
x=(1+3)/6 or (1-3)/6
x=4/6 or -2/6
x=2/3 or -1/3