Question

If \[f(x) = 4{x^2} + 3x - 7\] and \[\alpha \] is a common root of the equation \[{x^2} - 3x + 2 = 0\] and \[{x^2} + 2x - 3 = 0\] then find the value of \[f\left( \alpha \right)\].

Answer 1

Hint:
According to the question, find the roots of \[{x^2} - 3x + 2 = 0\] and \[{x^2} + 2x - 3 = 0\] by splitting the middle term method. Then take the common from the roots of the equation and substitute that value in \[f(x) = 4{x^2} + 3x - 7\].

Complete step by step solution:
Firstly we will take the equation, \[{x^2} - 3x + 2 = 0\]
Now, we will calculate the roots by using splitting the middle term method.
In this method we will find out two numbers whose sum is \[ - 3\] and product is \[2\].
So, we are getting with the two numbers which are \[ - 2\] and \[ - 1\]
\[ \Rightarrow {x^2} - x - 2x + 2 = 0\]
Taking out common in the pairs of 2 we get,
\[ \Rightarrow x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0\]
Taking 2 same factors one time we get,
\[ \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) = 0\]
Now, we will separate the above two factors to calculate the value of x.
Firstly we will take the factor\[x - 1 = 0\]
Taking 1 on the right side we get,
Therefore, \[x = 1\]
Secondly we will take the factor \[x - 2 = 0\]
Taking 2 on the right side we get,
Therefore, \[x = 2\]
Hence, the value of \[x = 1,2\]
Then we will take the equation, \[{x^2} + 2x - 3 = 0\]
Now, we will calculate the roots by using splitting the middle term method.
In this method we will find out two numbers whose sum is 2 and product is \[ - 3\].
So, we are getting with the two numbers which are \[ - 1\] and \[3\]
\[ \Rightarrow {x^2} - x + 3x - 3 = 0\]
Taking out common in the pairs of 2 we get,
\[ \Rightarrow x\left( {x - 1} \right) + 3\left( {x - 1} \right) = 0\]
Taking 2 same factors one time we get,
\[ \Rightarrow \left( {x - 1} \right)\left( {x + 3} \right) = 0\]
Now, we will separate the above two factors to calculate the value of p.
Firstly we will take the factor\[x - 1 = 0\]
Taking 1 on the right side we get,
Therefore, \[x = 1\]
Secondly we will take the factor \[x + 3 = 0\]
Taking 3 on the right side we get,
Therefore, \[x = - 3\]
Hence, the value of \[x = 1, - 3\]
As it is given in the question that \[{x^2} - 3x + 2 = 0\] and \[{x^2} + 2x - 3 = 0\] and there is a common root that is \[\alpha \] .
But, it is clear from the calculated roots that \[\alpha = 1\] (As 1 is common root in both the equations)
So, we will calculate \[f\left( \alpha \right) = f(1)\] by substituting 1 in the given equation that is \[f(x) = 4{x^2} + 3x - 7\]
After substituting we get,
\[f(1) = 4{\left( 1 \right)^2} + 3 \times 1 - 7\]
On simplifying we get,
\[f(1) = 7 - 7 = 0\]

Hence, the value of \[f\left( \alpha \right) = f\left( 1 \right) = 0\]

Note:
To, solve these types of questions, you firstly need to find the roots of the given quadratic equations with splitting the middle term method or by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] as the equation is in the form of \[a{x^2} + bx + c = 0\] .Hence, put the common value in the given function and simplify it to get the desired result.