Solve 4x/10+2/5x=1

$$\frac { 4 x } { 10 } + \frac { 2 } { 5 x } = 1$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of 10,5x.

$$x\times 4x+2\times 2=10x$$

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

$$x^{2}\times 4+2\times 2=10x$$

Multiply 2 and 2 to get 4.

$$x^{2}\times 4+4=10x$$

Subtract 10x from both sides.

$$x^{2}\times 4+4-10x=0$$

Divide both sides by 2.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=-5 ab=2\times 2=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 2x^{2}-5x+2 as \left(2x^{2}-4x\right)+\left(-x+2\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(2x-1\right)$$

To find equation solutions, solve x-2=0 and 2x-1=0.

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

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 10x, the least common multiple of 10,5x.

$$x\times 4x+2\times 2=10x$$

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

$$x^{2}\times 4+2\times 2=10x$$

Multiply 2 and 2 to get 4.

$$x^{2}\times 4+4=10x$$

Subtract 10x from both sides.

$$x^{2}\times 4+4-10x=0$$

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.

$$4x^{2}-10x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -10 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$x=\frac{-\left(-10\right)±\sqrt{100-64}}{2\times 4}$$

Add 100 to -64.

$$x=\frac{-\left(-10\right)±\sqrt{36}}{2\times 4}$$

Take the square root of 36.

$$x=\frac{-\left(-10\right)±6}{2\times 4}$$

The opposite of -10 is 10.

$$x=\frac{10±6}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{10±6}{8}$$

Now solve the equation x=\frac{10±6}{8} when ± is plus. Add 10 to 6.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{10±6}{8} when ± is minus. Subtract 6 from 10.

$$x=\frac{4}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

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 10x, the least common multiple of 10,5x.

$$x\times 4x+2\times 2=10x$$

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

$$x^{2}\times 4+2\times 2=10x$$

Multiply 2 and 2 to get 4.

$$x^{2}\times 4+4=10x$$

Subtract 10x from both sides.

$$x^{2}\times 4+4-10x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}\times 4-10x=-4$$

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.

$$4x^{2}-10x=-4$$

Divide both sides by 4.

$$\frac{4x^{2}-10x}{4}=-\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{10}{4}\right)x=-\frac{4}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

Divide -4 by 4.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Add -1 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{3}{4} x-\frac{5}{4}=-\frac{3}{4}$$

Add \frac{5}{4} to both sides of the equation.

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