Question

How do you solve \[2{x^2} - 2x - 1 = 0\] ?

Answer 1

Hint: The given equation is a quadratic equation. Here, in this question to solve or find a roots by different methods firstly by using the method of factorisation otherwise you can also solve this by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . By further simplification, you get the required result.

Complete step-by-step answer:
Factorization means the process of creating a list of factors or roots otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.
The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\] . Here, b is the sum of d and e and c is the product of d and e, and further equating each term to zero.
Consider the given expression
 \[ \Rightarrow 2{x^2} - 2x - 1 = 0\] .
Here the coefficient of \[{x^2}\] is not unity, it is hard to find root by factorization.
Hence, solve this problem by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . The standard quadratic equation is in the form \[a{x^2} + bx + c = 0\]
 Compare the given equation \[2{x^2} - 2x - 1 = 0\] to the standard form of the quadratic equation
Here a=2, b=-2 and c=-1 then substituting the values to the formula, then
 \[ \Rightarrow x = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 2 \right)\left( { - 1} \right)} }}{{2\times 2}}\]
On simplification we have
 \[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 - \left( { - 8} \right)} }}{4}\]
 \[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 8} }}{4}\]
 \[ \Rightarrow x = \dfrac{{2 \pm \sqrt {12} }}{4}\]
 \[\sqrt {12} \] Can be written as \[\sqrt {12} = \sqrt {4 \times 3} = \sqrt 4 \times \sqrt 3 = \sqrt {{2^2}} \cdot \sqrt 3 = 2\sqrt 3 \]
 \[ \Rightarrow x = \dfrac{{2 \pm 2\sqrt 3 }}{4}\]
Take 2 as common in numerator
 \[\, \Rightarrow x = \dfrac{{2\left( {1 \pm \sqrt 3 } \right)}}{4}\]
 \[\, \Rightarrow x = \dfrac{{1 \pm \sqrt 3 }}{2}\]
Let write two terms separately, then
 \[\, \Rightarrow x = \dfrac{{1 + \sqrt 3 }}{2}\] or \[x = \dfrac{{1 - \sqrt 3 }}{2}\]
Hence, after solving the given quadratic equation \[2{x^2} - 2x - 1 = 0\] we get the roots of \[x = \dfrac{{1 + \sqrt 3 }}{2}\] or \[x = \dfrac{{1 - \sqrt 3 }}{2}\] .

So, the correct answer is “ \[x = \dfrac{{1 + \sqrt 3 }}{2}\] or \[x = \dfrac{{1 - \sqrt 3 }}{2}\] ”.

Note: The quadratic equation can be solved by using the factorisation method and we also find the roots by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . While factorising we use sum product rule, the sum product rule is given as the product factors of the number c is equal to the sum of the factors which satisfies the value of b