The following is a student s solution for solving $2 x ^ { 2 } = 1 - 5 x$ . What error did the student make?
$\begin{array} { l }
2 x ^ { 2 } = 1 - 5 x \
2 x ^ { 2 } - 1 + 5 x = 0 \
x = \frac { 1 \pm \sqrt { ( - 1 ) ^ { 2 } - 4 ( 2 ) ( 5 ) } } { 4 } \
x = \frac { 1 \pm \sqrt { 1 - 40 } } { 4 } \
x = \frac { 1 \pm \sqrt { - 39 } } { 4 } \
x = \frac { 1 } { 4 } \pm \frac { \sqrt { 39 } } { 4 } i
\end{array}$

Question

The following is a student s solution for solving $2 x ^ { 2 } = 1 - 5 x$ . What error did the student make?
 $\begin{array} { l } 
2 x ^ { 2 } = 1 - 5 x \
2 x ^ { 2 } - 1 + 5 x = 0 \
x = \frac { 1 \pm \sqrt { ( - 1 ) ^ { 2 } - 4 ( 2 ) ( 5 ) } } { 4 } \
x = \frac { 1 \pm \sqrt { 1 - 40 } } { 4 } \
x = \frac { 1 \pm \sqrt { - 39 } } { 4 } \
x = \frac { 1 } { 4 } \pm \frac { \sqrt { 39 } } { 4 } i
\end{array}$

Quizplus · Accepted Answer

The Answer of The following is a student s solution for solving $2 x ^ { 2 } = 1 - 5 x$ . What error did the student make?
 $\begin{array} { l } 
2 x ^ { 2 } = 1 - 5 x \
2 x ^ { 2 } - 1 + 5 x = 0 \
x = \frac { 1 \pm \sqrt { ( - 1 ) ^ { 2 } - 4 ( 2 ) ( 5 ) } } { 4 } \
x = \frac { 1 \pm \sqrt { 1 - 40 } } { 4 } \
x = \frac { 1 \pm \sqrt { - 39 } } { 4 } \
x = \frac { 1 } { 4 } \pm \frac { \sqrt { 39 } } { 4 } i
\end{array}$

The Following Is a Student S Solution for Solving 2x2=1−5x2 x ^ { 2 } = 1 - 5 x2x2=1−5x

The Following Is a Student S Solution for Solving $2 x ^ { 2 } = 1 - 5 x$