Solve (x)=16x^2-40x+25

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

Solve for x

$$x = \frac{25}{16} = 1\frac{9}{16} = 1.5625$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 16x^{2} from both sides.

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

Add 40x to both sides.

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

Combine x and 40x to get 41x.

$$41x-16x^{2}=25$$

Subtract 25 from both sides.

$$41x-16x^{2}-25=0$$

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

$$-16x^{2}+41x-25=0$$

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.

$$a+b=41 ab=-16\left(-25\right)=400$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 41.

$$a=25 b=16$$

Rewrite -16x^{2}+41x-25 as \left(-16x^{2}+25x\right)+\left(16x-25\right).

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

Factor out -x in -16x^{2}+25x.

$$-x\left(16x-25\right)+16x-25$$

Factor out common term 16x-25 by using distributive property.

$$\left(16x-25\right)\left(-x+1\right)$$

To find equation solutions, solve 16x-25=0 and -x+1=0.

$$x=\frac{25}{16} x=1$$

Steps Using the Quadratic Formula

Subtract 16x^{2} from both sides.

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

Add 40x to both sides.

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

Combine x and 40x to get 41x.

$$41x-16x^{2}=25$$

Subtract 25 from both sides.

$$41x-16x^{2}-25=0$$

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.

$$-16x^{2}+41x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 41 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-41±\sqrt{41^{2}-4\left(-16\right)\left(-25\right)}}{2\left(-16\right)}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\left(-16\right)\left(-25\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-41±\sqrt{1681+64\left(-25\right)}}{2\left(-16\right)}$$

Multiply 64 times -25.

$$x=\frac{-41±\sqrt{1681-1600}}{2\left(-16\right)}$$

Add 1681 to -1600.

$$x=\frac{-41±\sqrt{81}}{2\left(-16\right)}$$

Take the square root of 81.

$$x=\frac{-41±9}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-41±9}{-32}$$

Now solve the equation x=\frac{-41±9}{-32} when ± is plus. Add -41 to 9.

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

Divide -32 by -32.

$$x=1$$

Now solve the equation x=\frac{-41±9}{-32} when ± is minus. Subtract 9 from -41.

$$x=-\frac{50}{-32}$$

Reduce the fraction \frac{-50}{-32} to lowest terms by extracting and canceling out 2.

$$x=\frac{25}{16}$$

The equation is now solved.

$$x=1 x=\frac{25}{16}$$

Steps for Completing the Square

Subtract 16x^{2} from both sides.

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

Add 40x to both sides.

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

Combine x and 40x to get 41x.

$$41x-16x^{2}=25$$

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.

$$-16x^{2}+41x=25$$

Divide both sides by -16.

$$\frac{-16x^{2}+41x}{-16}=\frac{25}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\frac{41}{-16}x=\frac{25}{-16}$$

Divide 41 by -16.

$$x^{2}-\frac{41}{16}x=\frac{25}{-16}$$

Divide 25 by -16.

$$x^{2}-\frac{41}{16}x=-\frac{25}{16}$$

Divide -\frac{41}{16}, the coefficient of the x term, by 2 to get -\frac{41}{32}. Then add the square of -\frac{41}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{41}{16}x+\left(-\frac{41}{32}\right)^{2}=-\frac{25}{16}+\left(-\frac{41}{32}\right)^{2}$$

Square -\frac{41}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{41}{16}x+\frac{1681}{1024}=-\frac{25}{16}+\frac{1681}{1024}$$

Add -\frac{25}{16} to \frac{1681}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{41}{16}x+\frac{1681}{1024}=\frac{81}{1024}$$

Factor x^{2}-\frac{41}{16}x+\frac{1681}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{32}\right)^{2}=\frac{81}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{32}\right)^{2}}=\sqrt{\frac{81}{1024}}$$

Simplify.

$$x-\frac{41}{32}=\frac{9}{32} x-\frac{41}{32}=-\frac{9}{32}$$

Add \frac{41}{32} to both sides of the equation.

$$x=\frac{25}{16} x=1$$