Question

What is are the solutions to the statement:(2x-7)(4x^2+14x+49)=0

Answer 1

SOLUTION

Given the question, the following are the solution steps to answer the question.

STEP 1: Write the given equation.

$(2x-7)(4x^2+14x+49)=0$

STEP 2: Find the solutions to the statement

First splitting the equation into two, we have:

$\begin{gathered} (2x-7)(4x^{2}+14x+49)=0 \\ 2x-7=0 \\ 4x^2+14x+49=0 \\ \\ 2x-7=0 \\ 2x=7 \\ x=\frac{7}{2} \end{gathered}$

STEP 3: Find the solution to the second equation

$\begin{gathered} 4x^2+14x+49=0 \\ Using\text{ quadratic formula:} \\ x_{1,\:2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ From\text{ the equation,} \\ a=4,b=14,c=49 \\ \\ By\text{ substitution,} \\ x_{1,\:2}=\frac{-14\pm \sqrt{14^2-4\cdot \:4\cdot \:49}}{2\cdot \:4} \\ By\text{ simplifying the numerator,} \\ \sqrt{14^2-4\times4\times49}=\sqrt{196-784}=\sqrt{-588}=14\sqrt{3}i \end{gathered}$

By substitution,

$\begin{gathered} x_{1,\:2}=\frac{-14\pm \:14\sqrt{3}i}{2\cdot \:4} \\ \mathrm{Separate\:the\:solutions} \\ x_1=\frac{-14+14\sqrt{3}i}{2\cdot \:4},\:x_2=\frac{-14-14\sqrt{3}i}{2\cdot \:4} \\ x_1=\frac{-14+14\sqrt{3}i}{2\cdot\:4}=-\frac{7}{4}+i\frac{7\sqrt{3}}{4} \\ x_2=\frac{-14-14\sqrt{3}i}{2\cdot\:4}=-\frac{7}{4}-i\frac{7\sqrt{3}}{4} \end{gathered}$

Hence, the solutions to the statement are:

$\frac{7}{2},-\frac{7}{4}+i\frac{7\sqrt{3}}{4},-\frac{7}{4}-i\frac{7\sqrt{3}}{4}$