Solve 4x^2-9x+26=8x+8

$$4 x ^ { 2 } - 9 x + 26 = 8 x + 8$$

Solve for x

$$x=2$$

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Steps Using Factoring By Grouping

Subtract 8x from both sides.

$$4x^{2}-9x+26-8x=8$$

Combine -9x and -8x to get -17x.

$$4x^{2}-17x+26=8$$

Subtract 8 from both sides.

$$4x^{2}-17x+26-8=0$$

Subtract 8 from 26 to get 18.

$$4x^{2}-17x+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+18. To find a and b, set up a system to be solved.

$$a+b=-17 ab=4\times 18=72$$

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 72.

$$-1,-72 -2,-36 -3,-24 -4,-18 -6,-12 -8,-9$$

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -17.

$$a=-9 b=-8$$

Rewrite 4x^{2}-17x+18 as \left(4x^{2}-9x\right)+\left(-8x+18\right).

$$\left(4x^{2}-9x\right)+\left(-8x+18\right)$$

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

$$x\left(4x-9\right)-2\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(x-2\right)$$

To find equation solutions, solve 4x-9=0 and x-2=0.

$$x=\frac{9}{4} x=2$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$4x^{2}-9x+26-8x=8$$

Combine -9x and -8x to get -17x.

$$4x^{2}-17x+26=8$$

Subtract 8 from both sides.

$$4x^{2}-17x+26-8=0$$

Subtract 8 from 26 to get 18.

$$4x^{2}-17x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -17 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 4\times 18}}{2\times 4}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 4\times 18}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-17\right)±\sqrt{289-16\times 18}}{2\times 4}$$

Multiply -16 times 18.

$$x=\frac{-\left(-17\right)±\sqrt{289-288}}{2\times 4}$$

Add 289 to -288.

$$x=\frac{-\left(-17\right)±\sqrt{1}}{2\times 4}$$

Take the square root of 1.

$$x=\frac{-\left(-17\right)±1}{2\times 4}$$

The opposite of -17 is 17.

$$x=\frac{17±1}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{17±1}{8}$$

Now solve the equation x=\frac{17±1}{8} when ± is plus. Add 17 to 1.

$$x=\frac{18}{8}$$

Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{17±1}{8} when ± is minus. Subtract 1 from 17.

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

Divide 16 by 8.

$$x=2$$

The equation is now solved.

$$x=\frac{9}{4} x=2$$

Steps for Completing the Square

Subtract 8x from both sides.

$$4x^{2}-9x+26-8x=8$$

Combine -9x and -8x to get -17x.

$$4x^{2}-17x+26=8$$

Subtract 26 from both sides.

$$4x^{2}-17x=8-26$$

Subtract 26 from 8 to get -18.

$$4x^{2}-17x=-18$$

Divide both sides by 4.

$$\frac{4x^{2}-17x}{4}=-\frac{18}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{17}{4}x=-\frac{18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{17}{4}x=-\frac{9}{2}$$

Divide -\frac{17}{4}, the coefficient of the x term, by 2 to get -\frac{17}{8}. Then add the square of -\frac{17}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{4}x+\left(-\frac{17}{8}\right)^{2}=-\frac{9}{2}+\left(-\frac{17}{8}\right)^{2}$$

Square -\frac{17}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{4}x+\frac{289}{64}=-\frac{9}{2}+\frac{289}{64}$$

Add -\frac{9}{2} to \frac{289}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{4}x+\frac{289}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{17}{4}x+\frac{289}{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}.

$$\left(x-\frac{17}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x-\frac{17}{8}=\frac{1}{8} x-\frac{17}{8}=-\frac{1}{8}$$

Add \frac{17}{8} to both sides of the equation.

$$x=\frac{9}{4} x=2$$