Question

question 8 of 10 which shows the equation below written in standard form? 11 - 4x=(3x - 2)^2+1 a 9x^2 - 16x - 6 = 0 b 9x^2 - 8x + 16 = 0 c 9x^2 - 16x + 16 = 0 d 9x^2 - 8x - 6 = 0

Explanation:

Step1: Expand the right - hand side

Expand $(3x - 2)^2$ using $(a - b)^2=a^{2}-2ab + b^{2}$. Here $a = 3x$ and $b = 2$, so $(3x - 2)^2=(3x)^{2}-2\times3x\times2 + 2^{2}=9x^{2}-12x + 4$. The original equation $11-4x=(3x - 2)^2+1$ becomes $11-4x=9x^{2}-12x + 4+1$.

Step2: Simplify the right - hand side

Combine like terms on the right - hand side: $11-4x=9x^{2}-12x + 5$.

Step3: Rearrange to standard form

Move all terms to one side of the equation to get the standard quadratic form $ax^{2}+bx + c = 0$. Add $12x$ to both sides: $11-4x+12x=9x^{2}-12x + 5+12x$, which simplifies to $11 + 8x=9x^{2}+5$. Then subtract $11 + 8x$ from both sides: $0=9x^{2}-8x - 6$.

Answer:

D. $9x^{2}-8x - 6 = 0$