Affected Quadratic Equations.

Since $(x+b{)}^2=x^2+2bx+b^2,\text{and}(x-b{)}^2=x^2-2bx+b^2,$ it is evident that the expression $x^2+2bx$ or $x^2-2bx$ lacks only the third term, $b^2$, of being a perfect square.

This third term is the square of half the coefficient of $x$.

Every affected quadratic may be made to assume the form $x^2+2bx=c$ or $x^2-2bx=c$, by dividing the equation through by the coefficient of $x^2$.

To solve such an equation:

The first step is to add to both members the square of half the coefficient of $x$. This is called completing the square.

The second step is to extract the square root of each member of the resulting equation.

The third step is to reduce the two resulting simple equations.

1. Solve the equation $x^2-8x=20$.

lint We have x^2 - 8x = 20. lint Complete the square, x^2 - 8x + 16 = 36. lint Extract the square root, x - 4 = ±6. [1] lint Reduce, using the upper sign, x = 4 + 6 = 10, lintor using the lower sign, x = 4 - 6 = -2.

The roots are $10$ and $-2$.

Verify by putting these numbers for $x$ in the given equation. $$\begin{gathered} rcl<|>rclx\&=\&10,\&x\&=\&-2, \\ {10}^2-8(10)\&=\&20,\&(-2{)}^2-8(-2)\&=\&20, \\ 100-80\&=\&20.\&4+16\&=\&20. \end{gathered}$$

2. Solve the equation ${\displaystyle \frac{x+1}{x-1}}={\displaystyle \frac{4x-3}{x+9}}$.

lint Free from fractions, (x + 1)(x + 9) = (x - 1)(4x - 3). lint Therefore, -3x^2 + 17x = -6.

Since the square root of a negative number cannot be taken, the coefficient of $x^2$ must be changed to $+$.

lint Divide by $-3$, x^2 - 173x = 2. [1] Half the coefficient of $x$ is $\frac{1}{2}$ of $-\frac{17}{3}=-\frac{17}{6}$, and the square of $-\frac{17}{6}$ is $\frac{289}{36}$. Add the square of $-\frac{17}{6}$ to both sides, and we have x^2 - 17x3 + (176)^2 = 2 + 28936. [1] lint Now 2 + 28936 = 7236 + 28936 = 36136, linttherefore, x^2 - 173x + (176)^2 = 36136. [1] lint Extract the root, x - 176 = ±196. [1] lint Reduce, x - 176 = ±196. x = 176 + 196 = 366 = 6, lintor x = 176 - 196 = -26 = -13.

The roots are $6$ and $-{\displaystyle \frac{1}{3}}$.

Verify by putting these numbers for $x$ in the original equation. $[t]{rcl<{\qquad}|} x &=& 6. \ \dfrac{6 + 1}{6 - 1} &=& \dfrac{24 - 3}{6 + 9}. \ \dfrac{7}{5} &=& \dfrac{21}{15} \ \dfrac{7}{5} &=& \dfrac{7}{5} \

[t]{>{\qquad}rcl} x &=& -\dfrac{1}{3} \ \dfrac{-\dfrac{1}{3} + 1}{-\dfrac{1}{3} - 1} &=& \dfrac{-\dfrac{4}{3} - 3}{-\dfrac{1}{3} + 9}. \ -\dfrac{2}{4} &=& -\dfrac{13}{26}.$