Question

4m 2+ m - 17 = 0 how to solve step by step

Answer 1

The solutions for this quadratic equation are
`m_1 = 1.94, m_2 = -2.19`. Bhaskara was used to solve.

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\Delta))/(2*a)`

`x_(2) = (-b - √(\Delta))/(2*a)`

`\Delta = b^(2) - 4ac`

In this question:

We have the quadratic equation:

`4m^2 + m - 17 = 0`, which has
`a = 4, b = 1, c = -17`.

Then

`\Delta = (1)^(2) - 4(4)(-17) = 273`

`m_(1) = (-1 + √(273))/(8) = 1.94`

`m_(2) = (-1 - √(273))/(8) = -2.19`

The solutions for this quadratic equation are
`m_1 = 1.94, m_2 = -2.19`. Bhaskara was used to solve.