Question

Solve 3x - 5x+2; the Quadratic
Equation in completing Square
method​

Answer 1

$\large\bf\underline{Given:-}$

• quadratic equation :- 3x² -5x + 2

$\large\bf\underline {To \: find:-}$

• Value of x by completing the square method.

$\huge\bf\underline{Solution:-}$

$\rm \dashrightarrow \: 3 {x}^{2} - 5x + 2 \\ \\ \rm \dag \: divide \: the \: equation \: by \: 3 \\ \\ \rm \dashrightarrow \: \: \frac{ 3{x}^{2} }{3} - \frac{5x}{3} + \frac{2}{3} \\ \\ \rm \dashrightarrow \: {x}^{2} - \frac{ 5x}{3} + \frac{2}{3} = 0\\ \\ \rm \dag \: add \: (\frac{5}{6} ) {}^{2} \: on \: both \: sides \: of \: the \: equations \\ \\ \rm \dashrightarrow \: {x}^{2} - \frac{5x}{3} + \frac{25}{36} = \frac{ - 2}{3} + \frac{25}{36} \\ \\ \rm \dashrightarrow \: (x + \frac{5}{6} ) {}^{2} = - \frac{2}{3} + \frac{25}{36} \\ \\ \rm \dashrightarrow \: (x + \frac{5}{6} ) {}^{2} = \frac{ - 24 + 25}{36} \\ \\ \rm \dashrightarrow \:(x + \frac{5}{6} ) {}^{2} = \frac{1}{36} \\ \\ \rm \dashrightarrow \:x + \frac{5}{6} = \pm \sqrt{ \frac{1}{36} } \\ \\ \rm \dashrightarrow \:x + \frac{5}{6} = \pm\frac{1}{6} \\ \\ \rm \dashrightarrow \:x = \frac{1}{6} - \frac{5}{6} \\ \\ \rm \dashrightarrow \:x = \cancel\frac{ - 4}{6} \\ \\ \bf \dashrightarrow \:x = \frac{ - 2}{3} \\ \\ \rm \dashrightarrow \:or \: x = \frac{ - 1}{6} - \frac{5}{6} \\ \\ \rm \dashrightarrow \:x = \frac{ - 6}{6} \\ \\ \bf \dashrightarrow \:x = - 1$

$\large\bigstar \bf \: x = - 1 \: or \: x = \frac{ - 2}{3}$

Answer 2

$\bf\large{\underline{Question:-}}$

Solve 3x - 5x+2; the Quadratic Equation in completing Square method.

$\bf\large{\underline{Given:-}}$

• 3x-5x+2

$\bf\large{\underline{To\:solve:-}}$

• 3x-5x+2 using completing square method.

$\bf\large{\underline{Solution:-}}$

• 1st we divide the equation by 3 And after that we add (5/6)^2 in both side.

$\bf\large{\underline{According\:to\:Question:-}}$

$\tt→ \frac{3x^2}{3}-\frac{5x}{3}+\frac{2}{3}=0\\\tt→ x^2-\frac{5x}{3}+\frac{2}{3}=0\\\tt→ x^2-\frac{5x}{3}+(\frac{5}{6})^2=\frac{-2}{3}+(\frac{5}{6})^2\\\tt→ (x+\frac{5}{6})^2=\frac{-2}{3}+\frac{25}{36}\\\tt→ (x+\frac{5}{6})^2=\frac{-24+25}{36}\\\tt→(x+\frac{5}{6})^2=\frac{1}{36}\\\tt→ (x+\frac{5}{5})=±\sqrt\frac{1}{36}\\\tt→ x+\frac{5}{6}=±\frac{1}{6}\\\tt→ x=\frac{1}{6}-\frac{5}{6}\\\tt→ x=\frac{-4}{6}\\\tt→ x=\frac{-2}{3} \\\tt→ or\: x=\frac{-1}{6}-\frac{5}{6}\\\tt→ x=\frac{-6}{6}\\\tt→ x=-1$

Hence,

$\bf\large{\underline{\fbox{x=\frac{-2}{3}}}}$

Or,

$\bf\large{\underline{\fbox{x=-1}}}$