which constant must be added and subtracted to solve the quadratic equation 9 X square + 3 by 4 x minus root 2 is equal to zero
Answers
Answer:
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Answer:
Given quadratic equation:
9 x^{2}+\frac{3}{4} x-2=09x
2
+
4
3
x−2=0
(3 x)^{2}-(2 \times 3 x) \times\left(\frac{1}{8}\right)-2=0(3x)
2
−(2×3x)×(
8
1
)−2=0
By adding and subtracting, \left(\frac{1}{8}\right)^{2}(
8
1
)
2
on the above equation,
(3 x)^{2}-(2 \times 3 x) \times\left(\frac{1}{8}\right)+\left(\frac{1}{8}\right)^{2}-\left(\frac{1}{8}\right)^{2}=2 \rightarrow(i)(3x)
2
−(2×3x)×(
8
1
)+(
8
1
)
2
−(
8
1
)
2
=2→(i)
By using the formula, (a-b)^{2}(a−b)
2
\left(3 x-\frac{1}{8}\right)^{2}+\left(\frac{1}{8}\right)^{2}=2(3x−
8
1
)
2
+(
8
1
)
2
=2
\left(3 x-\frac{1}{8}\right)^{2} = 2-\left(\frac{1}{8}\right)^{2}(3x−
8
1
)
2
=2−(
8
1
)
2
\left(3 x-\frac{1}{8}\right)^{2} = 2-\left(\frac{1}{64}\right)(3x−
8
1
)
2
=2−(
64
1
)
\left(3 x-\frac{1}{8}\right)^{2} = \frac{128-1}{64}(3x−
8
1
)
2
=
64
128−1
\left(3 x-\frac{1}{8}\right)^{2} = \frac{127}{64}(3x−
8
1
)
2
=
64
127
From (i)
Here we added and subtracted
\left(\frac{1}{8}\right)^{2} = \frac{1}{64}(
8
1
)
2
=
64
1
is used to resolve the quadratic equation by the completing square.