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Question
question 3 of 10 which shows the equation below written in the form ax^2 + bx + c = 0? x + 10 = 3(x - 1)^2 a. 3x^2 - 5x - 7 = 0 b. 3x^2 - 5x + 13 = 0 c. 3x^2 - 7x - 7 = 0 d. 3x^2 - 7x + 13 = 0
Step1: Expand the right - hand side
Expand $3(x - 1)^2$ using the formula $(a - b)^2=a^{2}-2ab + b^{2}$. Here $a = x$ and $b = 1$, so $3(x - 1)^2=3(x^{2}-2x + 1)=3x^{2}-6x+3$.
Step2: Rearrange the equation
The original equation $x + 10=3(x - 1)^2$ becomes $x + 10=3x^{2}-6x + 3$. Move all terms to one side to get $0=3x^{2}-6x - x+3 - 10$.
Step3: Combine like terms
Combine the $x$ - terms and the constant terms: $3x^{2}+(-6x - x)+(3 - 10)=3x^{2}-7x - 7$. So the equation in the form $ax^{2}+bx + c = 0$ is $3x^{2}-7x - 7 = 0$.
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C. $3x^{2}-7x - 7 = 0$