Question

Find the roots of 35x^2 + 13x - 12 = 0

Answer 1

$35 {x}^{2} + 13x - 12 = 0 \\ 35 {x}^{2} + (28 - 15)x - 12 = 0 \\ 35 {x}^{2} + 28x - 15x - 12 = 0\\ 7x(5x + 4) - 3(5x + 4) = 0\\ (7x - 3)(5x + 4) = 0$

Answer 2

Answer:

The roots of 35x^2 + 13x - 12 = 0 are $- \frac{4}{5}$and $\frac{3}{7}$

Step-by-step explanation:

Here given,

$35 {x}^{2} + 13x - 12 = 0$

By middle term process we can solve this problem.

So,

$35 {x}^{2} + (28 - 15)x - 12 = 0 \\ 35 {x}^{2} + 28x - 15x - 12 = 0 \\ 7x(5x + 4) - 3(5x + 4) = 0 \\ (5x + 4)(7x - 3) = 0$

We know that if product of two terms is zero then they are separately zero.

$5x + 4 = 0 \\ 5x = - 4 \\ x = - \frac{4}{5}$

and

$7x - 3 = 0 \\ 7x = 3 \\ x = \frac{3}{7}$

So required roots of x are $- \frac{4}{5}$and $\frac{3}{7}$

This is a problem of Algebra.

Some important formulas of Algebra

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)