Question

Solve \[ - 2{x^2} + 5x - 2 = 0\]

Answer 1

Hint: According to the given question, the equation \[ - 2{x^2} + 5x - 2 = 0\] in the form of \[a{x^2} + bx + c = 0\] . Then, find out the product which is equal to ac and sum which is equal to b. Then use the splitting the middle term method to factorise the equation and then find the roots of the equation.

Complete step-by-step answer:
 As, it is given \[ - 2{x^2} + 5x - 2 = 0\]
Here, by using splitting the middle term method we can calculate the factors of the equation \[ - 2{x^2} + 5x - 2 = 0\]
In this method we will find out two numbers whose sum is \[5\] and product is \[4\] .
So, we are getting the two numbers which are \[4\] and \[1\] .
Now, solving with equation: \[ - 2{x^2} + 5x - 2 = 0\]
Splitting the \[5x\] into two numbers as \[4x + x\]
We get,
 \[ - 2{x^2} + 4x + x - 2 = 0\]
Taking out common in the pairs of 2 we get,
 \[ - 2x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0\]
Taking 2 same factors one time we get,
 \[\left( { - 2x + 1} \right)\left( {x - 2} \right) = 0\]
Now we will solve these two factors separately,
one factor as
 \[ - 2x + 1 = 0\]
Taking 1 from left hand side to right hand side we get,
 \[ - 2x = - 1\]
Cancelling negative side from both left and right hand side we get,
 \[2x = 1\]
Dividing both left and right hand side by 4 we get the value of x that is,
 \[x = \dfrac{1}{2}\]
Other factor as,
 \[x - 2 = 0\]
Taking 2 from left hand side to right hand side we get,
 \[x = 2\]
Hence, after solving the equation \[ - 2{x^2} + 5x - 2 = 0\] the value of \[x = 2,\dfrac{1}{2}\]
So, the correct answer is “\[x = 2,\dfrac{1}{2}\]”.

Note: To solve these types of questions we can either use the formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] in which the equation must be in the form of \[a{x^2} + bx + c = 0\] to find the roots of the equation or by using splitting the middle term method you can also calculate the factors if the equation is not given equal to zero as well as the value of x or roots of the equation as shown above.