Tell whether the equation has one,zero or many solutions -(2x+2)-1=-x-(x+3)

Mathematics

1 answer:

PtichkaEL [24]2 years ago

3 0

- (2x + 2) - 1 = -x - (x + 3)

-2x - 2 - 1 = -x - x - 3

-2x - 3 = -2x - 3

This is an identity and x can be any value

Answer many solutions

You might be interested in

Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).

KonstantinChe [14]

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

$2x^2-3x-6=0$

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

$2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}$

We have to find the value of (a+2)(b+2).

Putting the values:

$(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4$