#### Question

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.

#### Solution

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.