Question

How do you solve \[{x^2} - 2x = - 2\] ?

Answer 1

Hint: Given is a quadratic equation only needed to take the constant term on the left hand side. Then we can factorize the middle term or can use a quadratic formula. Factorizing middle term is nothing but finding a pair of numbers such that their product is last term and their addition is the second term. But here we will use quadratic formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
Given that \[{x^2} - 2x = - 2\]
Taking constant on left hand side,
 \[{x^2} - 2x + 2 = 0\]
Now we can see this equation is of the form \[a{x^2} + bx + c = 0\] that is a general quadratic equation. Now we will use quadratic formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
So comparing with the general equation we get a=1, b=-2 and c=2.
So we can write the equation above as,
 \[ = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 1 \times 2} }}{{2 \times 1}}\]
On solving the brackets and roots,
 \[ = \dfrac{{2 \pm \sqrt {4 - 8} }}{2}\]
 \[ = \dfrac{{2 \pm \sqrt { - 4} }}{2}\]
We can write the root as
 \[ = \dfrac{{2 \pm 2\sqrt { - 1} }}{2}\]
Taking 2 common we get,
 \[ = \dfrac{{2\left( {1 \pm \sqrt { - 1} } \right)}}{2}\]
Cancelling 2
 \[ = 1 \pm \sqrt { - 1} \]
We know that \[\sqrt { - 1} = i\]
 So substituting it we get the factors as \[1 + i \;and \;1 - i\] .
So, the correct answer is “ \[1 + i \;and \;1 - i\] ”.

Note: Note that if the value of discriminant \[\sqrt {{b^2} - 4ac} \] is negative we cannot ignore it. That is an indication that the roots are imaginary. So do observe the value of the discriminant. Also note that it is not necessary that every problem of quadratic equations can be solved by finding the factors. But it can be solved each time using quadratic formulas only for quadratic equations.