Question

question 6 of 10
which shows the equation below written in standard form?
13 - 6x = (2x - 5)² + 3

a. 4x² - 26x + 15 = 0
b. 4x² - 14x - 15 = 0
c. 4x² - 14x + 15 = 0
d. 4x² - 26x - 15 = 0

Explanation:

Step1: Expand the right - hand side

We know that \((a - b)^2=a^{2}-2ab + b^{2}\). For \((2x - 5)^2\), where \(a = 2x\) and \(b = 5\), we have \((2x-5)^{2}=(2x)^{2}-2\times(2x)\times5 + 5^{2}=4x^{2}-20x + 25\).
So the equation \(13-6x=(2x - 5)^{2}+3\) becomes \(13-6x=4x^{2}-20x + 25+3\).

Step2: Simplify the right - hand side

Simplify \(4x^{2}-20x + 25+3\) to get \(4x^{2}-20x+28\). The equation is now \(13-6x = 4x^{2}-20x + 28\).

Step3: Move all terms to one side

Subtract \(13\) and add \(6x\) to both sides of the equation to get \(0=4x^{2}-20x + 28-13 + 6x\).

Step4: Combine like terms

Combine the \(x\) terms: \(-20x+6x=-14x\), and combine the constant terms: \(28 - 13 = 15\). So the equation becomes \(4x^{2}-14x + 15=0\).

Answer:

C. \(4x^{2}-14x + 15=0\)