Question

Solve the quadratic equation $3{(2x - 1)^2} + 4(2x - 1) - 4 = 0$

Answer 1

Hint: Here we are asked to solve the given quadratic equation that is we have to find its roots. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.
The word quadratic means second degree values of the given variables.
The formula that we need to know before solving the problem:
Let \[a{x^2} + bx + c = 0\] be a quadratic equation then the roots of this equation are given by \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Since given that $3{(2x - 1)^2} + 4(2x - 1) - 4 = 0$ now let us convert the given into some form, by assuming that $t = 2x - 1$ and after solving using the quadratic formula, we will again convert them back to normal with the variable $x$
It is given that $3{(2x - 1)^2} + 4(2x - 1) - 4 = 0 \Rightarrow 3{(t)^2} + 4(t) - 4 = 0$ we aim to solve this equation that is we have to find its roots.
We know that the number of roots of an equation is equal to its degree. Here the degree of the given equation is two thus this equation will have two roots.
The roots of a quadratic equation can be found by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] where \[a\] - coefficient of the term \[{x^2}\], \[b\] - coefficient of the term \[x\], and \[c\] - constant term.
First, let us collect the required terms for the formula from the given quadratic equation to solve it.
From the given equation $3{(t)^2} + 4(t) - 4 = 0$, we have \[a = 3\], \[b = 4\], and \[c = - 4\].
On substituting these terms in the formula, we get
\[t = \dfrac{{ - \left( 4 \right) \pm \sqrt {{{\left( 4 \right)}^2} - 4\left( 3 \right)\left( { - 4} \right)} }}{{2\left( 3 \right)}}\]
On simplifying this we get
\[t = \dfrac{{ - 4 \pm \sqrt {16 + 48} }}{6}\]
On further simplification we get
\[t = \dfrac{{ - 4 \pm \sqrt {64} }}{6}\]
    \[ = \dfrac{{ - 4 \pm 8}}{6}\]
\[ \Rightarrow t = \dfrac{{ - 4 + 8}}{6}\] and \[ \Rightarrow t = \dfrac{{ - 4 - 8}}{6}\]
On solving the above, we get
\[ \Rightarrow t = \dfrac{2}{3}\] and \[ \Rightarrow t = - 2\]
Now again applying the value that $t = 2x - 1$ then we get \[ \Rightarrow 2x - 1 = \dfrac{2}{3}\] and \[ \Rightarrow 2x - 1 = - 2\]
Hence solving this we have, \[2x = \dfrac{2}{3} + 1 \Rightarrow x = \dfrac{5}{6}\] and \[ \Rightarrow 2x = - 2 + 1 \Rightarrow x = \dfrac{{ - 1}}{2}\]
Thus, we got the roots of the given quadratic equation that is \[x = \dfrac{5}{6}\] and \[x = \dfrac{{ - 1}}{2}\]


Note: The roots of the equation are nothing but the possible value of the unknown variable in that equation. Also, the number of roots of an equation depends on its degree. The degree of an equation is the highest power of the unknown variable in that equation.
Also, in the quadratic equation, the $a = 0$ is never possible, because then it will be a linear equation.
There are two more methods for solving quadratic equations:
1. Factorisation method.
2. Completing square method.