Question

Solve the polynomial $8{x^2} - 6x + 1 = 0$?

Answer 1

Hint: In this question we have to solve the given polynomial using quadratic formula. In the polynomial \[a{x^2} + bx + c\], where "\[a\]", "\[b\]", and “\[c\]" are real numbers and the Quadratic Formula is derived from the process of completing the square, and is formally stated as:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step answer:
The general form of quadratic equation is \[a{x^2} + bx + c = 0\], where "\[a\]", "\[b\]", and "\[c\]" are numerical coefficients or constant, and the value of is unknown. And one fundamental rule is that the value of \[a\], the first constant cannot be zero in a quadratic equation.
Now the given quadratic equation is,
$8{x^2} - 6x + 1 = 0$,
Now using the quadratic formula, which is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\],
Here\[a = 8\],\[b = - 6\],\[c = 1\],
Now substituting the values in the formula we get,
\[ \Rightarrow x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 8 \right)\left( 1 \right)} }}{{2\left( 8 \right)}}\],
Now simplifying we get,
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 - \left( {32} \right)} }}{{16}}\],
Now again simplifying we get,
\[ \Rightarrow x = \dfrac{{6 \pm \sqrt 4 }}{{16}}\],
Now taking the square root we get,
\[ \Rightarrow x = \dfrac{{6 \pm 2}}{{16}}\],
Now taking one value for $x$,i.e.,
$ \Rightarrow $\[x = \dfrac{{6 + 2}}{{16}} = \dfrac{8}{{16}} = \dfrac{1}{2}\],
Now taking other value for $x$ i.e.,
$ \Rightarrow x = \dfrac{{6 - 2}}{{16}} = \dfrac{4}{{16}} = \dfrac{1}{4}$,
Now we get two values of\[x\]they are \[x = \dfrac{1}{2}\]and\[x = \dfrac{1}{4}\].
So, by solving the quadratic equation we get,\[x = \dfrac{1}{2}\] and \[x = \dfrac{1}{4}\].
Final Answer:

\[\therefore \]If we solve the given equation, i.e., $8{x^2} - 6x + 1 = 0$, then the we get \[x = \dfrac{1}{2}\] and \[x = \dfrac{1}{4}\].

Note:
A linear equation is an equation of a straight line having a maximum of one variable. The degree of the variable will be equal to 1. To solve any equation in one variable, pit all the variable terms on the left hand side and all the numerical values on the right hand side to make the calculation solved easily.