Solve 25+x^2-21/10x=1/2

$$\frac { 25 + x ^ { 2 } - 21 } { 10 x } = \frac { 1 } { 2 }$$

Solve for x

$$x=1$$

$$x=4$$

Steps Using Factoring

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 10x,2.

$$25+x^{2}-21=5x$$

Subtract 21 from 25 to get 4.

$$4+x^{2}=5x$$

Subtract 5x from both sides.

$$4+x^{2}-5x=0$$

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

$$x^{2}-5x+4=0$$

To solve the equation, factor x^{2}-5x+4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-5 ab=4$$

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 factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-4\right)\left(x-1\right)$$

To find equation solutions, solve x-4=0 and x-1=0.

$$x=4 x=1$$

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 10x,2.

$$25+x^{2}-21=5x$$

Subtract 21 from 25 to get 4.

$$4+x^{2}=5x$$

Subtract 5x from both sides.

$$4+x^{2}-5x=0$$

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

$$x^{2}-5x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 4=4$$

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 x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).

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

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

$$x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-1\right)$$

To find equation solutions, solve x-4=0 and x-1=0.

$$x=4 x=1$$

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 10x,2.

$$25+x^{2}-21=5x$$

Subtract 21 from 25 to get 4.

$$4+x^{2}=5x$$

Subtract 5x from both sides.

$$4+x^{2}-5x=0$$

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.

$$x^{2}-5x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-\left(-5\right)±\sqrt{25-16}}{2}$$

Add 25 to -16.

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

Take the square root of 9.

$$x=\frac{-\left(-5\right)±3}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±3}{2}$$

Now solve the equation x=\frac{5±3}{2} when ± is plus. Add 5 to 3.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{5±3}{2} when ± is minus. Subtract 3 from 5.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=4 x=1$$

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 10x,2.

$$25+x^{2}-21=5x$$

Subtract 21 from 25 to get 4.

$$4+x^{2}=5x$$

Subtract 5x from both sides.

$$4+x^{2}-5x=0$$

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

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

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}$$

Add -4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{9}{4}$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=4 x=1$$