(4x-1)^2=(x-1)(x+1)-এর সমাধান করুন

$$( 4 x - 1 ) ^ { 2 } = ( x - 1 ) ( x + 1 )$$

Solve for x (complex solution)

$$x=\frac{4+\sqrt{14}i}{15}\approx 0.266666667+0.249443826i$$

$$x=\frac{-\sqrt{14}i+4}{15}\approx 0.266666667-0.249443826i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$16x^{2}-8x+1=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+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.

$$16x^{2}-8x+1=x^{2}-1$$

Subtract x^{2} from both sides.

$$16x^{2}-8x+1-x^{2}=-1$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-8x+1=-1$$

Add 1 to both sides.

$$15x^{2}-8x+1+1=0$$

Add 1 and 1 to get 2.

$$15x^{2}-8x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15\times 2}}{2\times 15}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 15\times 2}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60\times 2}}{2\times 15}$$

Multiply -60 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-120}}{2\times 15}$$

Add 64 to -120.

$$x=\frac{-\left(-8\right)±\sqrt{-56}}{2\times 15}$$

Take the square root of -56.

$$x=\frac{-\left(-8\right)±2\sqrt{14}i}{2\times 15}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{14}i}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{8±2\sqrt{14}i}{30}$$

Now solve the equation x=\frac{8±2\sqrt{14}i}{30} when ± is plus. Add 8 to 2i\sqrt{14}.

$$x=\frac{8+2\sqrt{14}i}{30}$$

Divide 8+2i\sqrt{14} by 30.

$$x=\frac{4+\sqrt{14}i}{15}$$

Now solve the equation x=\frac{8±2\sqrt{14}i}{30} when ± is minus. Subtract 2i\sqrt{14} from 8.

$$x=\frac{-2\sqrt{14}i+8}{30}$$

Divide 8-2i\sqrt{14} by 30.

$$x=\frac{-\sqrt{14}i+4}{15}$$

The equation is now solved.

$$x=\frac{4+\sqrt{14}i}{15} x=\frac{-\sqrt{14}i+4}{15}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$16x^{2}-8x+1=\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+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.

$$16x^{2}-8x+1=x^{2}-1$$

Subtract x^{2} from both sides.

$$16x^{2}-8x+1-x^{2}=-1$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-8x+1=-1$$

Subtract 1 from both sides.

$$15x^{2}-8x=-1-1$$

Subtract 1 from -1 to get -2.

$$15x^{2}-8x=-2$$

Divide both sides by 15.

$$\frac{15x^{2}-8x}{15}=-\frac{2}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}-\frac{8}{15}x=-\frac{2}{15}$$

Divide -\frac{8}{15}, the coefficient of the x term, by 2 to get -\frac{4}{15}. Then add the square of -\frac{4}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{15}x+\left(-\frac{4}{15}\right)^{2}=-\frac{2}{15}+\left(-\frac{4}{15}\right)^{2}$$

Square -\frac{4}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{15}x+\frac{16}{225}=-\frac{2}{15}+\frac{16}{225}$$

Add -\frac{2}{15} to \frac{16}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{15}x+\frac{16}{225}=-\frac{14}{225}$$

Factor x^{2}-\frac{8}{15}x+\frac{16}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{15}\right)^{2}=-\frac{14}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{15}\right)^{2}}=\sqrt{-\frac{14}{225}}$$

Simplify.

$$x-\frac{4}{15}=\frac{\sqrt{14}i}{15} x-\frac{4}{15}=-\frac{\sqrt{14}i}{15}$$

Add \frac{4}{15} to both sides of the equation.

$$x=\frac{4+\sqrt{14}i}{15} x=\frac{-\sqrt{14}i+4}{15}$$