Solve 16x^2+25=40x

$$16 x ^ { 2 } + 25 = 40 x$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using Factoring By Grouping

Subtract 40x from both sides.

$$16x^{2}+25-40x=0$$

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

$$16x^{2}-40x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-40 ab=16\times 25=400$$

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 400.

$$-1,-400 -2,-200 -4,-100 -5,-80 -8,-50 -10,-40 -16,-25 -20,-20$$

Calculate the sum for each pair.

$$-1-400=-401 -2-200=-202 -4-100=-104 -5-80=-85 -8-50=-58 -10-40=-50 -16-25=-41 -20-20=-40$$

The solution is the pair that gives sum -40.

$$a=-20 b=-20$$

Rewrite 16x^{2}-40x+25 as \left(16x^{2}-20x\right)+\left(-20x+25\right).

$$\left(16x^{2}-20x\right)+\left(-20x+25\right)$$

Factor out 4x in the first and -5 in the second group.

$$4x\left(4x-5\right)-5\left(4x-5\right)$$

Factor out common term 4x-5 by using distributive property.

$$\left(4x-5\right)\left(4x-5\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve 4x-5=0.

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

Steps Using the Quadratic Formula

Subtract 40x from both sides.

$$16x^{2}+25-40x=0$$

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.

$$16x^{2}-40x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -40 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 16\times 25}}{2\times 16}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 16\times 25}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-40\right)±\sqrt{1600-64\times 25}}{2\times 16}$$

Multiply -64 times 25.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1600}}{2\times 16}$$

Add 1600 to -1600.

$$x=\frac{-\left(-40\right)±\sqrt{0}}{2\times 16}$$

Take the square root of 0.

$$x=-\frac{-40}{2\times 16}$$

The opposite of -40 is 40.

$$x=\frac{40}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{40}{32}$$

Reduce the fraction \frac{40}{32} to lowest terms by extracting and canceling out 8.

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

Steps for Completing the Square

Subtract 40x from both sides.

$$16x^{2}+25-40x=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}-40x=-25$$

Divide both sides by 16.

$$\frac{16x^{2}-40x}{16}=-\frac{25}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{40}{16}\right)x=-\frac{25}{16}$$

Reduce the fraction \frac{-40}{16} to lowest terms by extracting and canceling out 8.

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

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.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-\frac{25}{16}+\left(-\frac{5}{4}\right)^{2}$$

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

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

Add -\frac{25}{16} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=0$$

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}.

$$\left(x-\frac{5}{4}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{5}{4}=0 x-\frac{5}{4}=0$$

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

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

The equation is now solved. Solutions are the same.

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