Question

How do you do 56 through 58? Thanks to anyone who awnsers!

Answer 1

56. 44

57. 2

58. Image attached.

First, let's simplify some things. Let's make this a quadratic equation.

Moving the -1 (by adding 1 to both sides) gives us 5x^2 + 8x + 1 = 0. Note that this is the standard form of a quadratic, which is ax^2 + bx + c = 0.

Now, the discriminant. The discriminant is helpful because it tells us if we have 2 real solutions, 1 real solution, or if this equation is imaginary and thus impossible to solve using elementary techniques. If the discriminant is greater (>) than 0, then it has 2 solutions. If the discriminant equals (=) 0, it has 1 solution. If the discriminant is less than (<) 0, then it is imaginary.

The discriminant of a quadratic equation is defined as b^2 - 4ac (from the quadratic formula.) We just plug in b, a and c. b will be 8, a will be 5, and c will be 1. Now we just solve. (8)^2 - 4 * 5 * 1 = 64 - 20 = 44. Since 44 is greater than 0, we have 2 solutions for this. 44 is the discriminant. (Answer to 56.) We also have 2 solutions (answer to 57.)

Now, I'll solve the quadratic equation using the quadratic formula. I'll attach my work in an image (answer to 58.)

Hope this helped!

Answer 2

Hello !

56)

$5 {x}^{2} + 8x = - 1 \\ 5 {x}^{2} + 8x + 1 = 0$

$\Delta = {b}^{2} - 4ac \\ = {8}^{2} - 4 \times 5 \times 1 \\ = 64 - 20 \\ = 44$

The value of the discriminant is 44.

57) ∆>0 : there are two solutions.

58)

$x_1 = \frac{ - b + \sqrt{\Delta} }{2a} \\ = \frac{ - 8 + \sqrt{44} }{2 \times 5} \\ = \frac{ \sqrt{11} - 4 }{5}$

$x_2 = \frac{ - b - \sqrt{\Delta} }{2a} \\ = \frac{ - 8 - \sqrt{44} }{2 \times 5} \\ = \frac{ - \sqrt{11} - 4 }{5}$

Have a nice day