Question

How do you solve $10{x^2} - 11x - 6 = 0$?

Answer 1

Hint: The above problem is based on the factorization of quadratic equations.
Quadratic equation is the one which has the degree of a variable as 2.
Quadratic equations will be factorized into factors by using the splitting the middle term method. As factorization is the method of breaking the term into its multiples or factors.

Complete step-by-step answer:
To factorize the above given quadratic equation we have two methods to solve one is the standard formula method and splitting the middle term, in this problem we will apply splitting the middle term method to factorize the quadratic equation.
In this method of factorization we have the equation of the form,
$a{x^2} + bx + c = 0$ , where a is the coefficient of x2, b is the coefficient of x and c is the constant.
Equation given to us;
$ \Rightarrow 10{x^2} - 11x - 6 = 0$
We have to find two numbers which on multiplication gives 60(60 because a and c are multiplied to give 60) and on addition or subtraction it gives 11.
Number of pair which gives the product as 60 are;
12 and 5, 10 and 6, 20 and 3, 30 and 2, 15 and 4.
Only 15 and 4 on subtraction will give 11.
$ \Rightarrow 10{x^2} - 15x + 4x - 6 = 0$(We have break the terms)
$ \Rightarrow 5x(2x - 3) + 2(2x - 3) = 0$ (Took out common from the given terms)
$ \Rightarrow (5x + 2)(2x - 3) = 0$

$ \Rightarrow x = \dfrac{{ - 2}}{5},x = \dfrac{3}{2}$ (Two factors of the equation)

Note:
There is another method of making the factors of quadratic equation which is the standard one;
$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a is the coefficient of $x_2$, b is the coefficient of x and c is the constant term. The above formula gives the two factors as the formula contains two signs plus and minus.