Q. 305 - Intermediate Algebra

Question

$18x^2-9=-21x$Solve the exercise

Step-by-Step Solution

Answer

The value of $x=\frac{1}{3} or x=\frac{-3}{2}$

Step 1. Given information

The given equation is$18x^2-9=-21x$

Step 2. Simplify the equation

$$\begin{gathered} \mathrm{Simplify} \mathrm{the} \mathrm{quadratic} \mathrm{expression} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{zero} \mathrm{product} \mathrm{property},\mathrm{that} \mathrm{is} \\ (\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})=0 \\ \mathrm{a}-\mathrm{b}=0 \mathrm{or} \mathrm{a}+\mathrm{b}=0 \\ \mathrm{a}=-\mathrm{b} \mathrm{or} \mathrm{a}={\mathrm{b}}^{} \\ \mathrm{Simplify} \mathrm{the} \mathrm{quadratic} \mathrm{expression}: \\ 18{\mathrm{x}}^2-9=-21\mathrm{x} [\mathrm{Given} \mathrm{the} \mathrm{expression}] \\ 18{\mathrm{x}}^2+21\mathrm{x}-9=0 [\mathrm{Write} \mathrm{the} \mathrm{quadratic} \mathrm{equation} \mathrm{of} \mathrm{standard} \mathrm{form}] \\ 3(6{\mathrm{x}}^2+7\mathrm{x}-3)=0,\left[\mathrm{Factorout}3\mathrm{fromexpression}\right] \\ (6{\mathrm{x}}^2+9\mathrm{x}-2\mathrm{x}-3)=0 [\mathrm{Since};7\mathrm{x}=9\mathrm{x}-2\mathrm{x}] \\ 3(3\mathrm{x}*(2\mathrm{x}+3)-1(2\mathrm{x}+3\left)\right)=0 [\mathrm{Factor} \mathrm{out} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{group}] \\ 3(3\mathrm{x}-1)*(2\mathrm{x}+3)=0 [\mathrm{Factor} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{expression}] \\ 3*0\mathrm{or}3\mathrm{x}-1=0 \mathrm{or} 2\mathrm{x}+3=0[\mathrm{Use} \mathrm{the} \mathrm{zero} \mathrm{product} \mathrm{property}] \\ or x-1/3 or x=-3/2 \end{gathered}$$

Step 3. Checking the equation

$$\begin{gathered} \mathrm{Check}: \mathrm{Substitute} \mathrm{the} \mathrm{value} :\mathrm{x}=\frac{1}{3} \mathrm{In} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 18x^2-9=-21x [\mathrm{Given} \mathrm{equation}] \\ 18(\frac{1}{3}{)}^2-9=-21\frac{3}{2}[\mathrm{Put} \mathrm{x}=\frac{1}{3}] \\ 2-9=-7 \\ -7=-7True \\ \mathrm{Substitute} \mathrm{the} \mathrm{value} \mathrm{x}=-\frac{3}{2} \mathrm{in} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 18x^2-9=-21x [\mathrm{Given} \mathrm{equation}\left] \\ 18\right(-\frac{3}{2}{)}^2-9=-21\frac{-3}{2}[Put x=-\frac{3}{2}] \\ \frac{81}{2}-9=\frac{63}{2} \\ \frac{63}{2}=\frac{63}{2} \\ \mathrm{Hence} \mathrm{the} \mathrm{answer} \mathrm{is} x=-\frac{3}{2} and x=\frac{1}{3} \end{gathered}$$