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day 2 - 9.9 & 9.10 - quadratic formula student name: 1. solve the follo…

Question

day 2 - 9.9 & 9.10 - quadratic formula
student name:

  1. solve the following equation:

0 = -2x² - 4x - 2

provide an answer accurate to the nearest hundredth.

  1. solve the following equation:

0 = 2x² + 8x + 6

provide an answer accurate to the nearest hundredth.

  1. solve the following equation:

-x² + 2x + 2 = 0

provide an answer accurate to the nearest hundredth.

Explanation:

Response
Problem 1:

Step1: Identify coefficients

For equation \(0 = -2x^2 - 4x - 2\), rewrite as \(-2x^2 - 4x - 2 = 0\). Here, \(a = -2\), \(b = -4\), \(c = -2\).

Step2: Apply quadratic formula

Quadratic formula: \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substitute values:
\(x=\frac{-(-4)\pm\sqrt{(-4)^2 - 4(-2)(-2)}}{2(-2)}=\frac{4\pm\sqrt{16 - 16}}{-4}=\frac{4\pm0}{-4}\).

Step3: Simplify

\(x=\frac{4}{-4}=-1\). So both roots are \(x = -1.00\) (to nearest hundredth).

Step1: Identify coefficients

For \(0 = 2x^2 + 8x + 6\), \(a = 2\), \(b = 8\), \(c = 6\).

Step2: Apply quadratic formula

\(x=\frac{-8\pm\sqrt{8^2 - 4(2)(6)}}{2(2)}=\frac{-8\pm\sqrt{64 - 48}}{4}=\frac{-8\pm\sqrt{16}}{4}=\frac{-8\pm4}{4}\).

Step3: Find two roots

First root: \(\frac{-8 + 4}{4}=\frac{-4}{4}=-1.00\). Second root: \(\frac{-8 - 4}{4}=\frac{-12}{4}=-3.00\).

Step1: Identify coefficients

For \(-x^2 + 2x + 2 = 0\), multiply by \(-1\) to get \(x^2 - 2x - 2 = 0\). So \(a = 1\), \(b = -2\), \(c = -2\).

Step2: Apply quadratic formula

\(x=\frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}=\frac{2\pm\sqrt{4 + 8}}{2}=\frac{2\pm\sqrt{12}}{2}=\frac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}\).

Step3: Calculate decimal values

\(\sqrt{3}\approx1.732\). So \(1 + 1.732\approx2.73\), \(1 - 1.732\approx -0.73\) (to nearest hundredth).

Answer:

\(x = -1.00\)

Problem 2: