Solve 3x^2+ax+7=3(x-2)^2+b

$$3 x ^ { 2 } + a x + 7 = 3 ( x - 2 ) ^ { 2 } + b$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{5+b-12x}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&x=0\text{ and }b=-5\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}+ax+7=3\left(x^{2}-4x+4\right)+b$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}+ax+7=3x^{2}-12x+12+b$$

Subtract 3x^{2} from both sides.

$$ax+7=3x^{2}-12x+12+b-3x^{2}$$

Combine 3x^{2} and -3x^{2} to get 0.

$$ax+7=-12x+12+b$$

Subtract 7 from both sides.

$$ax=-12x+12+b-7$$

Subtract 7 from 12 to get 5.

$$ax=-12x+5+b$$

The equation is in standard form.

$$xa=5+b-12x$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{5+b-12x}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{5+b-12x}{x}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{5+b-12x}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }b=-5\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}+ax+7=3\left(x^{2}-4x+4\right)+b$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}+ax+7=3x^{2}-12x+12+b$$

Subtract 3x^{2} from both sides.

$$ax+7=3x^{2}-12x+12+b-3x^{2}$$

Combine 3x^{2} and -3x^{2} to get 0.

$$ax+7=-12x+12+b$$

Subtract 7 from both sides.

$$ax=-12x+12+b-7$$

Subtract 7 from 12 to get 5.

$$ax=-12x+5+b$$

The equation is in standard form.

$$xa=5+b-12x$$

Divide both sides by x.

$$\frac{xa}{x}=\frac{5+b-12x}{x}$$

Dividing by x undoes the multiplication by x.

$$a=\frac{5+b-12x}{x}$$

Solve for b

$$b=ax+12x-5$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}+ax+7=3\left(x^{2}-4x+4\right)+b$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}+ax+7=3x^{2}-12x+12+b$$

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-12x+12+b=3x^{2}+ax+7$$

Subtract 3x^{2} from both sides.

$$-12x+12+b=3x^{2}+ax+7-3x^{2}$$

Combine 3x^{2} and -3x^{2} to get 0.

$$-12x+12+b=ax+7$$

Add 12x to both sides.

$$12+b=ax+7+12x$$

Subtract 12 from both sides.

$$b=ax+7+12x-12$$

Subtract 12 from 7 to get -5.

$$b=ax-5+12x$$