What are the solutions of -x^2+4x=x-4

Mathematics

1 answer:

Stolb23 [73]3 years ago

8 0

Answer:

x = -1 and x = 4

Step-by-step explanation:

Step 1: Get everything to one side of the equation so it's set equal to zero..

-x² + 4x = x - 4 becomes...

-x² + 3x + 4 = 0 (subtract x and add 4 to both sides)

x² - 3x - 4 = 0 (divide both sides by -1, we want the x² term to be positive)

(x - 4)(x + 1) = 0 (factor)

x - 4 = 0, then x = 4

and

x + 1 = 0, then x = -1

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Write (8a-^3) -2/3 in simplest form

Anna71 [15]

$(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Solution:

Given that,

$(8a^{-3})^{\frac{-2}{3}$

We have to write in simplest form

Use the following law of exponent

$(a^m)^n = a^{mn}$

Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$