
If 1 is a root of the quadratic equation \[{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + ax - 2 = 0}}\] and the quadratic equation \[{\text{a}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 6x}}} \right){\text{ - b = 0}}\] has equal roots, then what will be the value of b ?
Answer
514.5k+ views
Hint: Let us first, find the value of a and then put the value of a in the condition of equal roots to get the value of b.
Complete step-by-step answer:
Let, \[{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + ax - 2 = 0}}\] (1)
And, \[{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + 6ax - b = 0}}\] (2)
As it is given in the question that 1 is the root of equation 1.
And we know the condition of the roots of the equation that if p will be the root of any polynomial equation then p must satisfy that polynomial equation.
So, putting the value of x equal to 1 in equation 1. We get,
\[{\text{3}}{\left( 1 \right)^{\text{2}}}{\text{ + a}}\left( 1 \right){\text{ - 2 = 0}}\]
3 + a – 2 = 1 + a = 0
So, a = -1
Now, we had to find the value of b using equation 2.
So, as we know that for any quadratic equation \[p{x^{\text{2}}}{\text{ + qx + r = 0}}\]. If it has equal roots, then its determinant must be equal to zero.
And determinant of this quadratic equation will be \[{{\text{q}}^{\text{2}}}{\text{ - 4pr}}\]
As equation 2 has equal roots.
So, determinant of equation 2 will also be zero.
So, \[{\left( {{\text{6a}}} \right)^{\text{2}}}{\text{ - 4a}}\left( {{\text{ - b}}} \right){\text{ = 0}}\] (3)
Now putting the value of a in equation 3. We get,
\[{\left( {{\text{6}}\left( {{\text{ - 1}}} \right)} \right)^{\text{2}}}{\text{ - 4}}\left( {{\text{ - 1}}} \right)\left( {{\text{ - b}}} \right){\text{ = 0}}\]
\[{\left( {{\text{ - 6}}} \right)^{\text{2}}}{\text{ - 4b = 0}}\]
36 = 4b
Therefore, b = 9
Hence, the value of b will be equal to 9.
Note: Whenever we came up with this type of problem first, we had to find the value of a by putting the value of the given root of the equation to that equation. And after that we had to put the determinant of another given equation equal to zero because if roots of any quadratic equation are equal then its determinant must be equal to zero. This will be the easiest and efficient way to find the value of b.
Complete step-by-step answer:
Let, \[{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + ax - 2 = 0}}\] (1)
And, \[{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + 6ax - b = 0}}\] (2)
As it is given in the question that 1 is the root of equation 1.
And we know the condition of the roots of the equation that if p will be the root of any polynomial equation then p must satisfy that polynomial equation.
So, putting the value of x equal to 1 in equation 1. We get,
\[{\text{3}}{\left( 1 \right)^{\text{2}}}{\text{ + a}}\left( 1 \right){\text{ - 2 = 0}}\]
3 + a – 2 = 1 + a = 0
So, a = -1
Now, we had to find the value of b using equation 2.
So, as we know that for any quadratic equation \[p{x^{\text{2}}}{\text{ + qx + r = 0}}\]. If it has equal roots, then its determinant must be equal to zero.
And determinant of this quadratic equation will be \[{{\text{q}}^{\text{2}}}{\text{ - 4pr}}\]
As equation 2 has equal roots.
So, determinant of equation 2 will also be zero.
So, \[{\left( {{\text{6a}}} \right)^{\text{2}}}{\text{ - 4a}}\left( {{\text{ - b}}} \right){\text{ = 0}}\] (3)
Now putting the value of a in equation 3. We get,
\[{\left( {{\text{6}}\left( {{\text{ - 1}}} \right)} \right)^{\text{2}}}{\text{ - 4}}\left( {{\text{ - 1}}} \right)\left( {{\text{ - b}}} \right){\text{ = 0}}\]
\[{\left( {{\text{ - 6}}} \right)^{\text{2}}}{\text{ - 4b = 0}}\]
36 = 4b
Therefore, b = 9
Hence, the value of b will be equal to 9.
Note: Whenever we came up with this type of problem first, we had to find the value of a by putting the value of the given root of the equation to that equation. And after that we had to put the determinant of another given equation equal to zero because if roots of any quadratic equation are equal then its determinant must be equal to zero. This will be the easiest and efficient way to find the value of b.
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