Solve 64x^2-16x-35=0

$$64 { x }^{ 2 } -16x-35=0$$

Solve for x

$$x=-\frac{5}{8}=-0.625$$

$$x=\frac{7}{8}=0.875$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 64x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-16 ab=64\left(-35\right)=-2240$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -2240.

$$1,-2240 2,-1120 4,-560 5,-448 7,-320 8,-280 10,-224 14,-160 16,-140 20,-112 28,-80 32,-70 35,-64 40,-56$$

Calculate the sum for each pair.

$$1-2240=-2239 2-1120=-1118 4-560=-556 5-448=-443 7-320=-313 8-280=-272 10-224=-214 14-160=-146 16-140=-124 20-112=-92 28-80=-52 32-70=-38 35-64=-29 40-56=-16$$

The solution is the pair that gives sum -16.

$$a=-56 b=40$$

Rewrite 64x^{2}-16x-35 as \left(64x^{2}-56x\right)+\left(40x-35\right).

$$\left(64x^{2}-56x\right)+\left(40x-35\right)$$

Factor out 8x in the first and 5 in the second group.

$$8x\left(8x-7\right)+5\left(8x-7\right)$$

Factor out common term 8x-7 by using distributive property.

$$\left(8x-7\right)\left(8x+5\right)$$

To find equation solutions, solve 8x-7=0 and 8x+5=0.

$$x=\frac{7}{8} x=-\frac{5}{8}$$

Steps Using the Quadratic Formula

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.

$$64x^{2}-16x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 64 for a, -16 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 64\left(-35\right)}}{2\times 64}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 64\left(-35\right)}}{2\times 64}$$

Multiply -4 times 64.

$$x=\frac{-\left(-16\right)±\sqrt{256-256\left(-35\right)}}{2\times 64}$$

Multiply -256 times -35.

$$x=\frac{-\left(-16\right)±\sqrt{256+8960}}{2\times 64}$$

Add 256 to 8960.

$$x=\frac{-\left(-16\right)±\sqrt{9216}}{2\times 64}$$

Take the square root of 9216.

$$x=\frac{-\left(-16\right)±96}{2\times 64}$$

The opposite of -16 is 16.

$$x=\frac{16±96}{2\times 64}$$

Multiply 2 times 64.

$$x=\frac{16±96}{128}$$

Now solve the equation x=\frac{16±96}{128} when ± is plus. Add 16 to 96.

$$x=\frac{112}{128}$$

Reduce the fraction \frac{112}{128} to lowest terms by extracting and canceling out 16.

$$x=\frac{7}{8}$$

Now solve the equation x=\frac{16±96}{128} when ± is minus. Subtract 96 from 16.

$$x=-\frac{80}{128}$$

Reduce the fraction \frac{-80}{128} to lowest terms by extracting and canceling out 16.

$$x=-\frac{5}{8}$$

The equation is now solved.

$$x=\frac{7}{8} x=-\frac{5}{8}$$

Steps for Completing the Square

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.

$$64x^{2}-16x-35=0$$

Add 35 to both sides of the equation.

$$64x^{2}-16x-35-\left(-35\right)=-\left(-35\right)$$

Subtracting -35 from itself leaves 0.

$$64x^{2}-16x=-\left(-35\right)$$

Subtract -35 from 0.

$$64x^{2}-16x=35$$

Divide both sides by 64.

$$\frac{64x^{2}-16x}{64}=\frac{35}{64}$$

Dividing by 64 undoes the multiplication by 64.

$$x^{2}+\left(-\frac{16}{64}\right)x=\frac{35}{64}$$

Reduce the fraction \frac{-16}{64} to lowest terms by extracting and canceling out 16.

$$x^{2}-\frac{1}{4}x=\frac{35}{64}$$

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\frac{35}{64}+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{35+1}{64}$$

Add \frac{35}{64} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{9}{16}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{1}{8}=\frac{3}{4} x-\frac{1}{8}=-\frac{3}{4}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{7}{8} x=-\frac{5}{8}$$