Solve 2x + 2/x = 3 হলে , x ^ 2 + 1/(x ^ 2) এর মান কত ? | 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

$$2 x + \frac { 2 } { x } = 3$$

Solve for x (complex solution)

$x=\frac{3+\sqrt{7}i}{4}\approx 0.75+0.661437828i$
$x=\frac{-\sqrt{7}i+3}{4}\approx 0.75-0.661437828i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$2xx+2=3x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$2x^{2}+2=3x$$

Subtract $3x$ from both sides.

$$2x^{2}+2-3x=0$$

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.

$$2x^{2}-3x+2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-3$ for $b$, and $2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\times 2}}{2\times 2}$$

Square $-3$.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 2}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\times 2}}{2\times 2}$$

Multiply $-8$ times $2$.

$$x=\frac{-\left(-3\right)±\sqrt{9-16}}{2\times 2}$$

Add $9$ to $-16$.

$$x=\frac{-\left(-3\right)±\sqrt{-7}}{2\times 2}$$

Take the square root of $-7$.

$$x=\frac{-\left(-3\right)±\sqrt{7}i}{2\times 2}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±\sqrt{7}i}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{3±\sqrt{7}i}{4}$$

Now solve the equation $x=\frac{3±\sqrt{7}i}{4}$ when $±$ is plus. Add $3$ to $i\sqrt{7}$.

$$x=\frac{3+\sqrt{7}i}{4}$$

Now solve the equation $x=\frac{3±\sqrt{7}i}{4}$ when $±$ is minus. Subtract $i\sqrt{7}$ from $3$.

$$x=\frac{-\sqrt{7}i+3}{4}$$

The equation is now solved.

$$x=\frac{3+\sqrt{7}i}{4}$$ $$x=\frac{-\sqrt{7}i+3}{4}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$2xx+2=3x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$2x^{2}+2=3x$$

Subtract $3x$ from both sides.

$$2x^{2}+2-3x=0$$

Subtract $2$ from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-3x=-2$$

Divide both sides by $2$.

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

Dividing by $2$ undoes the multiplication by $2$.

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

Divide $-2$ by $2$.

$$x^{2}-\frac{3}{2}x=-1$$

Divide $-\frac{3}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{4}$. Then add the square of $-\frac{3}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{3}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-1+\frac{9}{16}$$

Add $-1$ to $\frac{9}{16}$.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{7}{16}$$

Factor $x^{2}-\frac{3}{2}x+\frac{9}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{7}i}{4}$$ $$x-\frac{3}{4}=-\frac{\sqrt{7}i}{4}$$

Add $\frac{3}{4}$ to both sides of the equation.

$$x=\frac{3+\sqrt{7}i}{4}$$ $$x=\frac{-\sqrt{7}i+3}{4}$$