x^2-25x+625=0-এর সমাধান করুন

$$x ^ { 2 } - 25 x + 625 = 0$$

Solve for x (complex solution)

$$x=\frac{25+25\sqrt{3}i}{2}\approx 12.5+21.650635095i$$

$$x=\frac{-25\sqrt{3}i+25}{2}\approx 12.5-21.650635095i$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}-25x+625=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -25 for b, and 625 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 625}}{2}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 625}}{2}$$

Multiply -4 times 625.

$$x=\frac{-\left(-25\right)±\sqrt{625-2500}}{2}$$

Add 625 to -2500.

$$x=\frac{-\left(-25\right)±\sqrt{-1875}}{2}$$

Take the square root of -1875.

$$x=\frac{-\left(-25\right)±25\sqrt{3}i}{2}$$

The opposite of -25 is 25.

$$x=\frac{25±25\sqrt{3}i}{2}$$

Now solve the equation x=\frac{25±25\sqrt{3}i}{2} when ± is plus. Add 25 to 25i\sqrt{3}.

$$x=\frac{25+25\sqrt{3}i}{2}$$

Now solve the equation x=\frac{25±25\sqrt{3}i}{2} when ± is minus. Subtract 25i\sqrt{3} from 25.

$$x=\frac{-25\sqrt{3}i+25}{2}$$

The equation is now solved.

$$x=\frac{25+25\sqrt{3}i}{2} x=\frac{-25\sqrt{3}i+25}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-25x+625=0$$

Subtract 625 from both sides of the equation.

$$x^{2}-25x+625-625=-625$$

Subtracting 625 from itself leaves 0.

$$x^{2}-25x=-625$$

Divide -25, the coefficient of the x term, by 2 to get -\frac{25}{2}. Then add the square of -\frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=-625+\left(-\frac{25}{2}\right)^{2}$$

Square -\frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-25x+\frac{625}{4}=-625+\frac{625}{4}$$

Add -625 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=-\frac{1875}{4}$$

Factor x^{2}-25x+\frac{625}{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}.

$$\left(x-\frac{25}{2}\right)^{2}=-\frac{1875}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{-\frac{1875}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{25\sqrt{3}i}{2} x-\frac{25}{2}=-\frac{25\sqrt{3}i}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{25+25\sqrt{3}i}{2} x=\frac{-25\sqrt{3}i+25}{2}$$