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प्रश्न
If 1 is a root of the quadratic equation \[3 x^2 + ax - 2 = 0\] and the quadratic equation \[a( x^2 + 6x) - b = 0\] has equal roots, find the value of b.
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उत्तर
The given quadratic equation is \[3 x^2 + ax - 2 = 0\]
and one root is 1.
Then, it satisfies the given equation.
\[3 \left( 1 \right)^2 + a\left( 1 \right) - 2 = 0\]
\[ \Rightarrow 3 + a - 2 = 0\]
\[ \Rightarrow 1 + a = 0\]
\[ \Rightarrow a = - 1\]
The quadratic equation \[a( x^2 + 6x) - b = 0\],has equal roots.
Putting the value of a, we get
\[- 1\left( x^2 + 6x \right) - b = 0\]
\[ \Rightarrow x^2 + 6x + b = 0\]
Here,
\[A = 1, B = 6 \text { and } C = b\]
As we know that \[D = B^2 - 4AC\]
Putting the values of \[A = 1, B = 6 \text { and } C = b\].
\[D = \left( 6 \right)^2 - 4\left( 1 \right)\left( b \right)\]
\[ = 36 - 4b\]
The given equation will have real and equal roots, if D = 0
Thus,
\[36 - 4b = 0\]
\[\Rightarrow 4b = 36\]
\[ \Rightarrow b = 9\]
Therefore, the value of b is 9.
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