
Find the values of a and b for which $ x = \dfrac{3}{4} $ and $ x = - 2 $ are the roots of the equation $ a{x^2} + bx - 6 = 0 $
Answer
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Hint: First of all we will frame two equations by placing the values of “x” one by one in and simplify to get the equations in the terms of “a” and “b”. Then by using the elimination method we will find the values of “a” and “b”.
Complete step-by-step answer:
Take the given expression: $ a{x^2} + bx - 6 = 0 $
Place the given value for $ x = \dfrac{3}{4} $
$ a{\left( {\dfrac{3}{4}} \right)^2} + b\left( {\dfrac{3}{4}} \right) - 6 = 0 $
Simplify the above expression –
$ a{\left( {\dfrac{9}{{16}}} \right)^{}} + \left( {\dfrac{{3b}}{4}} \right) - 6 = 0 $
Take LCM (least common multiple) for the above expression –
$ 9a + 12b - 96 = 0 $ ….. (A)
Similarly, place the given value for $ x = - 2 $
$ a{\left( { - 2} \right)^2} + b\left( { - 2} \right) - 6 = 0 $
Simplify the above expression, remember the square of negative term gives the positive term and the product of positive and negative term gives negative term.
$ 4a - 2b - 6 = 0 $ …. (B)
Multiply the above expression with the number
$ 24a - 12b - 36 = 0 $ …… (C)
Add equations (A) and (C)
$ 9a + 12b - 96 + 24a - 12b - 36 = 0 $
Combine the like terms in the above expression –
$ \underline {9a + 24a} + 1\underline {2b - 12b} - \underline {36 - 96} = 0 $
Like terms with the same value and opposite signs cancels each other.
$ 33a - 132 = 0 $
Make the required term “a” the subject, when you move any term from one side to another then the sign of the term also changes.
$ 33a = 132 $
Term multiplicative one side if moved to the opposite side then it goes to the denominator.
$ a = \dfrac{{132}}{{33}} $
Simplify –
$ a = 4 $ ….. (C)
Place the above value in equation (B)
$ 4(4) - 2b - 6 = 0 $
Simplify –
$
16 - 2b - 6 = 0 \\
10 = 2b \\
b = 5 \;
$
Hence, the required values for a and b are $ 4 $ and $ 5 $ respectively.
Note: We have two methods to solve the linear equation. They are elimination and the substitution methods. Be careful about the sign convention while simplifying. When any term is moved from one side to another then the sign of the term also changes where positive term changes to negative and vice-versa.
Complete step-by-step answer:
Take the given expression: $ a{x^2} + bx - 6 = 0 $
Place the given value for $ x = \dfrac{3}{4} $
$ a{\left( {\dfrac{3}{4}} \right)^2} + b\left( {\dfrac{3}{4}} \right) - 6 = 0 $
Simplify the above expression –
$ a{\left( {\dfrac{9}{{16}}} \right)^{}} + \left( {\dfrac{{3b}}{4}} \right) - 6 = 0 $
Take LCM (least common multiple) for the above expression –
$ 9a + 12b - 96 = 0 $ ….. (A)
Similarly, place the given value for $ x = - 2 $
$ a{\left( { - 2} \right)^2} + b\left( { - 2} \right) - 6 = 0 $
Simplify the above expression, remember the square of negative term gives the positive term and the product of positive and negative term gives negative term.
$ 4a - 2b - 6 = 0 $ …. (B)
Multiply the above expression with the number
$ 24a - 12b - 36 = 0 $ …… (C)
Add equations (A) and (C)
$ 9a + 12b - 96 + 24a - 12b - 36 = 0 $
Combine the like terms in the above expression –
$ \underline {9a + 24a} + 1\underline {2b - 12b} - \underline {36 - 96} = 0 $
Like terms with the same value and opposite signs cancels each other.
$ 33a - 132 = 0 $
Make the required term “a” the subject, when you move any term from one side to another then the sign of the term also changes.
$ 33a = 132 $
Term multiplicative one side if moved to the opposite side then it goes to the denominator.
$ a = \dfrac{{132}}{{33}} $
Simplify –
$ a = 4 $ ….. (C)
Place the above value in equation (B)
$ 4(4) - 2b - 6 = 0 $
Simplify –
$
16 - 2b - 6 = 0 \\
10 = 2b \\
b = 5 \;
$
Hence, the required values for a and b are $ 4 $ and $ 5 $ respectively.
Note: We have two methods to solve the linear equation. They are elimination and the substitution methods. Be careful about the sign convention while simplifying. When any term is moved from one side to another then the sign of the term also changes where positive term changes to negative and vice-versa.
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