Question

How do you factor \[2{x^2} - 4x + 18 = 0\] ?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step-by-step answer:
Given, \[2{x^2} - 4x + 18 = 0\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\] , we have \[a = 2\] , \[b = - 4\] and \[c = 18\] .
We cannot solve this by factorization method that is the standard factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\] , which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\] is impossible.
So we use the quadratic formula
 \[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
 \[ \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(2)(18)} }}{{2(2)}}\]
 \[ = \dfrac{{ + 4 \pm \sqrt {16 - (8)(18)} }}{4}\]
 \[ = \dfrac{{ 4 \pm \sqrt {16 - 144} }}{4}\]
 \[ = \dfrac{{ 4 \pm \sqrt { - 128} }}{4}\]
We know that \[\sqrt { - 1} = i\]
 \[ = \dfrac{{ 4 \pm i\sqrt {128} }}{4}\]
We can write \[128 = 64 \times 2\]
 \[ = \dfrac{{ 4 \pm i\sqrt {64 \times 2} }}{4}\]
We know 64 is a perfect square,
 \[ = \dfrac{{ 4 \pm 8i\sqrt 2 }}{4}\]
 \[ = \dfrac{{4\left( { 1 \pm 2i\sqrt 2 } \right)}}{4}\]
 \[ = 1 \pm 2i\sqrt 2 \]
Hence the roots of \[2{x^2} - 4x + 18 = 0\] are \[ 1 + 2i\sqrt 2 \] and \[ 1 - 2i\sqrt 2 \]
Hence the factors of \[2{x^2} - 4x + 18 = 0\] are \[x - \left( { - 1 + 2i\sqrt 2 } \right)\] and \[x - \left( { - 1 - 2i\sqrt 2 } \right)\]
So, the correct answer is “\[x - \left( { 1 + 2i\sqrt 2 } \right)\] and \[x - \left( { 1 - 2i\sqrt 2 } \right)\] ”.

Note: These are the roots of the given polynomial of degree 2. In above, if we are unable to expand the middle term of the given equation into a sum of two numbers then we use a quadratic formula to solve the given problem. Quadratic formula and Sridhar’s formula are both the same. Careful in the calculation part.