Question

Solve the following quadratic equation:
\[4{{x}^{2}}-4x-3=0\]
(a) \[x=\dfrac{3}{2},\dfrac{-1}{2}\]
(b) \[x=\dfrac{-1}{2},\dfrac{-3}{2}\]
(c) \[x=\dfrac{3}{2},\dfrac{1}{2}\]
(d) \[x=\dfrac{1}{2},1\]

Answer 1

Hint: Consider the given quadratic equation \[4{{x}^{2}}-4x-3=0\] and write it as \[4{{x}^{2}}+2x-6x-3=0\] and then factorize as (2x + 1) (2x – 3) = 0 and finally get the value of x.

Complete step-by-step answer:
We are given a quadratic equation \[4{{x}^{2}}-4x-3=0\] and we have to solve to find the values of x. \[4{{x}^{2}}-4x-3=0\] is considered as a quadratic equation as quadratic equation any equation that can be rearranged in standard form as \[a{{x}^{2}}+bx+c=0.\]
Here, x represents an unknown value, a, b and c are known numbers where \[a\ne 0.\] Therefore, it becomes linear as no \[a{{x}^{2}}\] term is there. The numbers a, b and c are coefficients of the equation and may be distinguished by calling them respectively the quadratic coefficient, the linear coefficient, and the constant or the free term.
The values of y that satisfy the equation are called solutions of the equation and roots or zeros of the expression on its left-hand side. A quadratic equation has the utmost two solutions. If there is no real solution, one says it is double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation of the form \[a{{y}^{2}}+by+c=0\] can be factored as \[a\left( y-\alpha \right)\left( y-\beta \right)=0\] where \[\alpha \text{ and }\beta \] are solutions of x. The quadratic equation only contains powers of x that are non-negative integers and therefore it is a polynomial equation. In particular, it is a second-degree polynomial equation.
As the equation given is,
\[4{{x}^{2}}-4x-3=0\]
This can be written as
\[\Rightarrow 4{{x}^{2}}+2x-6x-3=0\]
This can be further factorized as,
\[\Rightarrow \left( 2x+1 \right)\left( 2x-3 \right)=0\]
So, for the equation, the values \[\dfrac{-1}{2}\text{ and }\dfrac{3}{2}\] satisfies the equation in place of x.
Hence, the correct option is (a).

Note: We can also solve it using a formula or by the method \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] for quadratic equation \[a{{x}^{2}}+bx+c=0.\]