(2r+3)²+(r+4)²=10 into the form of ax²+bc+c=0... How do you solve that?

Please help. the packet is due tonight

Zolol [24]

Answer:

[C] $\displaystyle \frac{-3}{250}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
Factoring
Functions
Function Notation
Conjugations

Calculus

Limits
Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$
Limit Property [Multiplied Constant]: $\displaystyle \lim_{x \to c} bf(x) = b \lim_{x \to c} f(x)$
Derivatives
Definition of a Derivative: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle g(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$\displaystyle f(x) = \frac{3}{\sqrt{x - 4}}$

$\displaystyle g(29)$

Step 2: Differentiate

Substitute in function [Function g(x)]: $\displaystyle g(x) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{x + h - 4}} - \frac{3}{\sqrt{x - 4}}}{h}$
Substitute in x [Function g(x)]: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{29 + h - 4}} - \frac{3}{\sqrt{29 - 4}}}{h}$
Simplify: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{25 + h}} - \frac{3}{5}}{h}$
Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15}{5\sqrt{25 + h}} - \frac{3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$
[Subtraction] Combine like terms: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{15 - 3\sqrt{25 + h}}{5\sqrt{25 + h}}}{h}$
Factor: $\displaystyle g(29) = \lim_{h \to 0} \frac{\frac{3(5 - \sqrt{25 + h})}{5\sqrt{25 + h}}}{h}$
Rewrite: $\displaystyle g(29) = \lim_{h \to 0} \frac{3(5 - \sqrt{25 + h})}{5h\sqrt{25 + h}}$
Rewrite [Limit Property - Multiplied Constant]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}}$
Root Conjugation: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{5 - \sqrt{25 + h}}{h\sqrt{25 + h}} \cdot \frac{5 + \sqrt{25 + h}}{5 + \sqrt{25 + h}}$
Multiply: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{5h\sqrt{25 + h} + h^2 + 25h}$
Factor: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-h}{h(5\sqrt{25 + h} + h + 25)}$
Simplify: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + h} + h + 25}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle g(29) = \frac{3}{5} \lim_{h \to 0} \frac{-1}{5\sqrt{25 + 0} + 0 + 25}$
Simplify: $\displaystyle g(29) = \frac{3}{5} \cdot \frac{-1}{50}$
Multiply: $\displaystyle g(29) = \frac{-3}{250}$