Solve Find the values of a, b and c (2x+3)^ 2 equiv a(x+3)^ 2 +bx+c | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for a (complex solution)

$\left\{\begin{matrix}a=-\frac{-4x^{2}+bx-12x+c-9}{\left(x+3\right)^{2}}\text{, }&x\neq -3\\a\in \mathrm{C}\text{, }&b=\frac{c}{3}-3\text{ and }x=-3\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x+3\right)^{2}+bx+c$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x^{2}+6x+9\right)+bx+c$$

Use the distributive property to multiply $a$ by $x^{2}+6x+9$.

$$4x^{2}+12x+9=ax^{2}+6ax+9a+bx+c$$

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

$$ax^{2}+6ax+9a+bx+c=4x^{2}+12x+9$$

Subtract $bx$ from both sides.

$$ax^{2}+6ax+9a+c=4x^{2}+12x+9-bx$$

Subtract $c$ from both sides.

$$ax^{2}+6ax+9a=4x^{2}+12x+9-bx-c$$

Combine all terms containing $a$.

$$\left(x^{2}+6x+9\right)a=4x^{2}+12x+9-bx-c$$

The equation is in standard form.

$$\left(x^{2}+6x+9\right)a=4x^{2}-bx+12x-c+9$$

Divide both sides by $x^{2}+6x+9$.

$$\frac{\left(x^{2}+6x+9\right)a}{x^{2}+6x+9}=\frac{-bx+\left(2x+3\right)^{2}-c}{x^{2}+6x+9}$$

Dividing by $x^{2}+6x+9$ undoes the multiplication by $x^{2}+6x+9$.

$$a=\frac{-bx+\left(2x+3\right)^{2}-c}{x^{2}+6x+9}$$

Divide $-bx-c+\left(2x+3\right)^{2}$ by $x^{2}+6x+9$.

$$a=\frac{4x^{2}-bx+12x-c+9}{\left(x+3\right)^{2}}$$

Solve for b (complex solution)

$\left\{\begin{matrix}b=\frac{9-9a-c+12x-6ax+4x^{2}-ax^{2}}{x}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&a=-\frac{c}{9}+1\text{ and }x=0\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x+3\right)^{2}+bx+c$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x^{2}+6x+9\right)+bx+c$$

Use the distributive property to multiply $a$ by $x^{2}+6x+9$.

$$4x^{2}+12x+9=ax^{2}+6ax+9a+bx+c$$

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

$$ax^{2}+6ax+9a+bx+c=4x^{2}+12x+9$$

Subtract $ax^{2}$ from both sides.

$$6ax+9a+bx+c=4x^{2}+12x+9-ax^{2}$$

Subtract $6ax$ from both sides.

$$9a+bx+c=4x^{2}+12x+9-ax^{2}-6ax$$

Subtract $9a$ from both sides.

$$bx+c=4x^{2}+12x+9-ax^{2}-6ax-9a$$

Subtract $c$ from both sides.

$$bx=4x^{2}+12x+9-ax^{2}-6ax-9a-c$$

Reorder the terms.

$$bx=-ax^{2}+4x^{2}-6ax+12x-9a-c+9$$

The equation is in standard form.

$$xb=9-9a-c+12x-6ax+4x^{2}-ax^{2}$$

Divide both sides by $x$.

$$\frac{xb}{x}=\frac{-ax^{2}+\left(2x+3\right)^{2}-6ax-c-9a}{x}$$

Dividing by $x$ undoes the multiplication by $x$.

$$b=\frac{-ax^{2}+\left(2x+3\right)^{2}-6ax-c-9a}{x}$$

Divide $-ax^{2}-6ax-9a-c+\left(2x+3\right)^{2}$ by $x$.

$$b=-ax+4x-6a+\frac{9-9a-c}{x}+12$$

Solve for a

$\left\{\begin{matrix}a=-\frac{-4x^{2}+bx-12x+c-9}{\left(x+3\right)^{2}}\text{, }&x\neq -3\\a\in \mathrm{R}\text{, }&b=\frac{c}{3}-3\text{ and }x=-3\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x+3\right)^{2}+bx+c$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x^{2}+6x+9\right)+bx+c$$

Use the distributive property to multiply $a$ by $x^{2}+6x+9$.

$$4x^{2}+12x+9=ax^{2}+6ax+9a+bx+c$$

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

$$ax^{2}+6ax+9a+bx+c=4x^{2}+12x+9$$

Subtract $bx$ from both sides.

$$ax^{2}+6ax+9a+c=4x^{2}+12x+9-bx$$

Subtract $c$ from both sides.

$$ax^{2}+6ax+9a=4x^{2}+12x+9-bx-c$$

Combine all terms containing $a$.

$$\left(x^{2}+6x+9\right)a=4x^{2}+12x+9-bx-c$$

The equation is in standard form.

$$\left(x^{2}+6x+9\right)a=4x^{2}-bx+12x-c+9$$

Divide both sides by $x^{2}+6x+9$.

$$\frac{\left(x^{2}+6x+9\right)a}{x^{2}+6x+9}=\frac{-bx+\left(2x+3\right)^{2}-c}{x^{2}+6x+9}$$

Dividing by $x^{2}+6x+9$ undoes the multiplication by $x^{2}+6x+9$.

$$a=\frac{-bx+\left(2x+3\right)^{2}-c}{x^{2}+6x+9}$$

Divide $-bx-c+\left(2x+3\right)^{2}$ by $x^{2}+6x+9$.

$$a=\frac{4x^{2}-bx+12x-c+9}{\left(x+3\right)^{2}}$$

Solve for b

$\left\{\begin{matrix}b=\frac{9-9a-c+12x-6ax+4x^{2}-ax^{2}}{x}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&a=-\frac{c}{9}+1\text{ and }x=0\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x+3\right)^{2}+bx+c$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+3\right)^{2}$.

$$4x^{2}+12x+9=a\left(x^{2}+6x+9\right)+bx+c$$

Use the distributive property to multiply $a$ by $x^{2}+6x+9$.

$$4x^{2}+12x+9=ax^{2}+6ax+9a+bx+c$$

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

$$ax^{2}+6ax+9a+bx+c=4x^{2}+12x+9$$

Subtract $ax^{2}$ from both sides.

$$6ax+9a+bx+c=4x^{2}+12x+9-ax^{2}$$

Subtract $6ax$ from both sides.

$$9a+bx+c=4x^{2}+12x+9-ax^{2}-6ax$$

Subtract $9a$ from both sides.

$$bx+c=4x^{2}+12x+9-ax^{2}-6ax-9a$$

Subtract $c$ from both sides.

$$bx=4x^{2}+12x+9-ax^{2}-6ax-9a-c$$

Reorder the terms.

$$bx=-ax^{2}+4x^{2}-6ax+12x-9a-c+9$$

The equation is in standard form.

$$xb=9-9a-c+12x-6ax+4x^{2}-ax^{2}$$

Divide both sides by $x$.

$$\frac{xb}{x}=\frac{-ax^{2}+\left(2x+3\right)^{2}-6ax-c-9a}{x}$$

Dividing by $x$ undoes the multiplication by $x$.

$$b=\frac{-ax^{2}+\left(2x+3\right)^{2}-6ax-c-9a}{x}$$

Divide $-ax^{2}-6ax-9a-c+\left(2x+3\right)^{2}$ by $x$.

$$b=-ax+4x-6a+\frac{9-9a-c}{x}+12$$