y=x^2+(a-3)x+a-4-এর সমাধান করুন

$$y = x ^ { 2 } + ( a - 3 ) x + a - 4$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{x^{2}-3x-y-4}{x+1}\text{, }&x\neq -1\\a\in \mathrm{C}\text{, }&y=0\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-3 by x.

$$y=x^{2}+ax-3x+a-4$$

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

$$x^{2}+ax-3x+a-4=y$$

Subtract x^{2} from both sides.

$$ax-3x+a-4=y-x^{2}$$

Add 3x to both sides.

$$ax+a-4=y-x^{2}+3x$$

Add 4 to both sides.

$$ax+a=y-x^{2}+3x+4$$

Combine all terms containing a.

$$\left(x+1\right)a=y-x^{2}+3x+4$$

The equation is in standard form.

$$\left(x+1\right)a=4+y+3x-x^{2}$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)a}{x+1}=\frac{4+y+3x-x^{2}}{x+1}$$

Dividing by x+1 undoes the multiplication by x+1.

$$a=\frac{4+y+3x-x^{2}}{x+1}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{x^{2}-3x-y-4}{x+1}\text{, }&x\neq -1\\a\in \mathrm{R}\text{, }&y=0\text{ and }x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-3 by x.

$$y=x^{2}+ax-3x+a-4$$

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

$$x^{2}+ax-3x+a-4=y$$

Subtract x^{2} from both sides.

$$ax-3x+a-4=y-x^{2}$$

Add 3x to both sides.

$$ax+a-4=y-x^{2}+3x$$

Add 4 to both sides.

$$ax+a=y-x^{2}+3x+4$$

Combine all terms containing a.

$$\left(x+1\right)a=y-x^{2}+3x+4$$

The equation is in standard form.

$$\left(x+1\right)a=4+y+3x-x^{2}$$

Divide both sides by x+1.

$$\frac{\left(x+1\right)a}{x+1}=\frac{4+y+3x-x^{2}}{x+1}$$

Dividing by x+1 undoes the multiplication by x+1.

$$a=\frac{4+y+3x-x^{2}}{x+1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{4y+a^{2}-10a+25}-a+3}{2}$$

$$x=\frac{-\sqrt{4y+a^{2}-10a+25}-a+3}{2}$$

Solve for x

$$x=\frac{\sqrt{4y+a^{2}-10a+25}-a+3}{2}$$

$$x=\frac{-\sqrt{4y+a^{2}-10a+25}-a+3}{2}\text{, }y\geq -\frac{\left(a-5\right)^{2}}{4}$$