Question

Solve the given quadratic equation:
$48{x^2} - 13x - 1 = 0$.

Answer 1

Here we have a quadratic equation, we can solve this using factorization method or by quadratic formula or by graphing method or by completing square method. We solve this using the factorization method. If we fails to split the middle term we use the quadratic formula, that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step-by-step solution:
Given, $48{x^2} - 13x - 1 = 0$.
Now consider the equation $48{x^2} - 13x - 1 = 0$. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$. We have $a = 48$, $b = - 13$ and $c = - 1$.
For factorization, the standard equation is rewritten as $a{x^2} + {b_1}x + {b_2}x + c = 0$ such that${b_1} \times {b_2} = ac$ and${b_1} + {b_2} = b$.
Here we can say that ${b_1} = - 16$ and ${b_2} = 3$. Because ${b_1} \times {b_2} = - 48$ $(a \times c)$ and ${b_1} + {b_2} = - 13(b)$.
Now we write $48{x^2} - 13x - 1 = 0$ as,
$\Rightarrow 48{x^2} - 16x + 3x - 1 = 0$
Taking ‘16x’ common in the first two terms and taking $1$ common in the remaining two terms we have,
$\Rightarrow 16x(3x - 1) + 1(3x - 1) = 0$
Again taking $(3x - 1)$ common we have,
$\Rightarrow (3x - 1)(16x + 1) = 0$.
By zero product principle we have,
$\Rightarrow 3x - 1 = 0$ and $16x + 1 = 0$
$\Rightarrow 3x = 1$ and $16x = - 1$
$\Rightarrow x = \dfrac{1}{3}$ and $x = - \dfrac{1}{{16}}$
This is the required solution.

Note: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. Suppose if we have a cubic equation, then we need to reduce it to a quadratic equation using syntactic division and reduce it to a quadratic equation. We can solve the quadratic equation easily.