Q. 303 - Intermediate Algebra

Question

Solve the following exercise

$15x^2-10x=40$

Step-by-Step Solution

Answer

The value of $x=-\frac{4}{3},x=2$.

Step 1. Given information

The given equation $15x^2-10x=40$.

Step 2. Simplify the quadratic equation

$$\begin{gathered} \mathrm{Simplify} \mathrm{the} \mathrm{quadratic} \mathrm{expression}: \\ 15\mathrm{x}²-10\mathrm{x}=40 [\mathrm{Given} \mathrm{the} \mathrm{expression}] \\ 15x^2-10x-40=0 [\mathrm{Write} \mathrm{the} \mathrm{quadratic} \mathrm{equation} \mathrm{of} \mathrm{standard} \mathrm{form}] \\ 5(3x^2-2x-8)=0[\mathrm{Factor} \mathrm{out} 3 \mathrm{from} \mathrm{expression}] \\ 5(31²-6x+4x-8)=0 [\mathrm{since};-6\mathrm{x}=-6.\mathrm{x}+4\mathrm{x}] \\ 5(3x*(x-2)+4(x-2\left)\right)=0 [\mathrm{Factor} \mathrm{out} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{group}] \\ 5(3x+4)*(x-2)=0 [\mathrm{Factor} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{expression}] \\ 5-0or3x+4=0 or x-2=0 [\mathrm{Use} \mathrm{the} \mathrm{zero} \mathrm{product} \mathrm{property}] \\ \mathrm{or} \mathrm{x}=-\frac{4}{3}\mathrm{or} \mathrm{x}=2 \end{gathered}$$

$$\begin{gathered} (a-b)(a+b)=0 \\ a-b=0 \mathrm{or} a+b=0 \\ a=b \mathrm{or} a=-b \end{gathered}$$

Step 3. Checking the roots

$$\begin{gathered} \mathrm{Check}:\mathrm{Substitute} \mathrm{the} \mathrm{value} \mathrm{x}=-4/3 \mathrm{in} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 15{\mathrm{x}}^2-10\mathrm{x}=4\left(\right[\mathrm{Given} \mathrm{equation}) \\ 15*(-\frac{4}{3}{)}^2-10(-\frac{4}{3})=40] [\mathrm{Put} x=-\frac{4}{3}] \\ \frac{80}{3}+\frac{40}{3}=40 \\ 40=40True \\ \mathrm{Substitute} \mathrm{the} \mathrm{value} \mathrm{x}=2 \mathrm{in} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 15x^2-10x=40 [\mathrm{Given} \mathrm{equation}] \\ 15\times (2{)}^2-10\left(2\right)=40 [\mathrm{Put} x=2] \\ 60-20=40 \\ 40=40 \\ \mathrm{Therefore},\mathrm{answer} \mathrm{is} x=-4/3,x=2 \end{gathered}$$