-x^2+4x-4=-x-এর সমাধান করুন

$$- x ^ { 2 } + 4 x - 4 = - x$$

Solve for x

$$x=1$$

$$x=4$$

Steps Using Factoring By Grouping

Add x to both sides.

$$-x^{2}+4x-4+x=0$$

Combine 4x and x to get 5x.

$$-x^{2}+5x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -x^{2}+5x-4 as \left(-x^{2}+4x\right)+\left(x-4\right).

$$\left(-x^{2}+4x\right)+\left(x-4\right)$$

Factor out -x in -x^{2}+4x.

$$-x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+1\right)$$

To find equation solutions, solve x-4=0 and -x+1=0.

$$x=4 x=1$$

Steps Using the Quadratic Formula

Add x to both sides.

$$-x^{2}+4x-4+x=0$$

Combine 4x and x to get 5x.

$$-x^{2}+5x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-5±\sqrt{25-16}}{2\left(-1\right)}$$

Add 25 to -16.

$$x=\frac{-5±\sqrt{9}}{2\left(-1\right)}$$

Take the square root of 9.

$$x=\frac{-5±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±3}{-2}$$

Now solve the equation x=\frac{-5±3}{-2} when ± is plus. Add -5 to 3.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-5±3}{-2} when ± is minus. Subtract 3 from -5.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=1 x=4$$

Steps for Completing the Square

Add x to both sides.

$$-x^{2}+4x-4+x=0$$

Combine 4x and x to get 5x.

$$-x^{2}+5x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-x^{2}+5x=4$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=\frac{4}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=\frac{4}{-1}$$

Divide 4 by -1.

$$x^{2}-5x=-4$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=4 x=1$$