Question

How do you solve \[3{x^2} - 2x - 2 = 0\] ?

Answer 1

Hint: Here we are given with a quadratic equation. To solve the quadratic equation is nothing but finding the factors of the equation. Using a quadratic formula is finding the roots with the help of discriminant. We will compare the given quadratic equation with the general quadratic equation of the form \[a{x^2} + bx + c = 0\] . So let’s start solving!
Formula used:
Quadratic formula: \[ \Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
Given that \[3{x^2} - 2x - 2 = 0\] is the equation given.
Now comparing it with the general equation \[a{x^2} + bx + c = 0\] we get \[a = 3,b = - 2\& c = - 2\] .
Putting these values in quadratic formula,
 \[ \Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
 \[ \Rightarrow \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 3 \times \left( { - 2} \right)} }}{{2 \times 3}}\]
Now on solving the brackets,
 \[ \Rightarrow \dfrac{{2 \pm \sqrt {4 + 24} }}{6}\]
 \[ \Rightarrow \dfrac{{2 \pm \sqrt {28} }}{6}\]
Solving the root,
 \[ \Rightarrow \dfrac{{2 \pm \sqrt {4 \times 7} }}{6}\]
4 is the perfect square of 2 so on solving the root,
 \[ \Rightarrow \dfrac{{2 \pm 2\sqrt 7 }}{6}\]
Taking 2 common from numerator,
 \[ \Rightarrow \dfrac{{2\left( {1 \pm \sqrt 7 } \right)}}{6}\]
On dividing we get,
 \[ \Rightarrow \dfrac{{1 \pm \sqrt 7 }}{3}\]
This is our answer or we can say the roots are \[ \Rightarrow \dfrac{{1 + \sqrt 7 }}{3}\] or \[ \Rightarrow \dfrac{{1 - \sqrt 7 }}{3}\]
So, the correct answer is “ \[ \dfrac{{1 + \sqrt 7 }}{3}\] or \[ \dfrac{{1 - \sqrt 7 }}{3}\] ”.

Note: Note that quadratic formula is used to find the roots of a given quadratic equation. Sometimes we can factorize the roots directly. But quadratic formulas can be used generally to find the roots of any quadratic equation. The value of discriminant is used to decide the type of roots so obtained such that roots are equal or different and are real or not.