1/2x-1-1/2x+1=1/4=-এর সমাধান করুন

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps by Finding Square Root

Variable x cannot be equal to any of the values -\frac{1}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 4\left(2x-1\right)\left(2x+1\right), the least common multiple of 2x-1,2x+1,4.

$$8x+4-\left(8x-4\right)=\left(2x-1\right)\left(2x+1\right)$$

To find the opposite of 8x-4, find the opposite of each term.

$$8x+4-8x+4=\left(2x-1\right)\left(2x+1\right)$$

Combine 8x and -8x to get 0.

$$4+4=\left(2x-1\right)\left(2x+1\right)$$

Add 4 and 4 to get 8.

$$8=\left(2x-1\right)\left(2x+1\right)$$

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$8=\left(2x\right)^{2}-1$$

Expand \left(2x\right)^{2}.

$$8=2^{2}x^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$8=4x^{2}-1$$

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

$$4x^{2}-1=8$$

Add 1 to both sides.

$$4x^{2}=8+1$$

Add 8 and 1 to get 9.

$$4x^{2}=9$$

Divide both sides by 4.

$$x^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{2},\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 4\left(2x-1\right)\left(2x+1\right), the least common multiple of 2x-1,2x+1,4.

$$8x+4-\left(8x-4\right)=\left(2x-1\right)\left(2x+1\right)$$

To find the opposite of 8x-4, find the opposite of each term.

$$8x+4-8x+4=\left(2x-1\right)\left(2x+1\right)$$

Combine 8x and -8x to get 0.

$$4+4=\left(2x-1\right)\left(2x+1\right)$$

Add 4 and 4 to get 8.

$$8=\left(2x-1\right)\left(2x+1\right)$$

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$8=\left(2x\right)^{2}-1$$

Expand \left(2x\right)^{2}.

$$8=2^{2}x^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$8=4x^{2}-1$$

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

$$4x^{2}-1=8$$

Subtract 8 from both sides.

$$4x^{2}-1-8=0$$

Subtract 8 from -1 to get -9.

$$4x^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-9\right)}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\left(-9\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-9\right)}}{2\times 4}$$

Multiply -16 times -9.

$$x=\frac{0±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±12}{8}$$

Now solve the equation x=\frac{0±12}{8} when ± is plus. Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{0±12}{8} when ± is minus. Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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