Question

If \[\left( {3x + a} \right)\left( {3x + b} \right) = ab\], then value of \[x\] is equal to
A. \[0,\dfrac{{ - \left( {a + b} \right)}}{3}\]
B. \[5,\dfrac{{\left( {a - b} \right)}}{2}\]
C. \[7,\dfrac{{ab}}{3}\]
D. \[ - 2,\dfrac{{ab}}{2}\]

Answer 1

Hint:We are given with an expression and from the given expression we are asked to find out the value of \[x\]. The expression can be simplified further, so try to simplify the expression and take it into a form such that you can find out the value of \[x\].

You can also see that all the options have two values for \[x\] so, while solving the problem remember that you must get two values of \[x\].

Complete step by step solution:
Given the expression \[\left( {3x + a} \right)\left( {3x + b} \right) = ab\]

Let us simplify the expression,
\[\left( {3x + a} \right)\left( {3x + b} \right) = ab\]
\[ \Rightarrow 9{x^2} + 3xb + 3xa + ab = ab\]
\[ \Rightarrow 9{x^2} + 3xb + 3xa = 0\]

We can see that there is \[x\] in all the terms so, we can take \[x\] common from the terms and then the equation will be,
\[x\left( {9x + 3b + 3a} \right) = 0\]

From the above equation, we observe that from the product of \[x\] and \[\left( {9x + 3b + 3a} \right)\] to be zero. One of the terms should be zero. That is,
\[x = 0\] or \[9x + 3b + 3a = 0\]

The term \[\left( {9x + 3b + 3a} \right)\] can be further simplified as,
\[3x + b + a = 0\]
\[ \Rightarrow 3x = - \left( {a + b} \right)\]
\[ \Rightarrow x = \dfrac{{ - \left( {a + b} \right)}}{3}\]

Therefore, the value of \[x\] can be either \[x = 0\] or \[x = \dfrac{{ - \left( {a + b} \right)}}{3}\].

Hence, the correct answer is option (A) \[0,\dfrac{{ - \left( {a + b} \right)}}{3}\]

Note:Here the equation formed after simplification is a quadratic equation. A quadratic equation is an equation of second degree. And the roots of a quadratic equation \[a{x^2} + bx + c = 0\] can be found out using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Using this formula too, the above problem can be solved. But as here we could take the term \[x\] common so it was easy for us to solve that way. In some problems you might not be able to take \[x\] common, in such cases you need to find the roots using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].