Question

Factories X square minus 3x minus 1

Answer 1

Answer:

${x}^{2} - 3x - 1 = 0$

$=>{x}^{2}-3x-1-{(\frac{3}{2}) }^{2}+{( \frac{3}{2}) }^{2}=0$

$=>{x}^{2}-2\times\frac{3}{2} x+(\frac{3}{2})^{2}-1- {(\frac{3}{2})}^{2}=0$

$=>{x}^{2}-2\times (\frac{3}{2})^{2}x+{(\frac{3}{2}) }^{2}= {(\frac{3}{2})}^{2}+1$

$= > {(x - \frac{3}{2} )}^{2} = \frac{9}{4} +1$

$= > \sqrt{ {(x - \frac{3}{2}) }^{2} } = \sqrt \frac{9 + 4}{4}$

$= > x - \frac{3}{2} = \sqrt \frac{13}{4}$

$=> x=\sqrt{\frac{13}{4}}+\frac{3}{2}=\sqrt{ \frac{13}{4} }+ \sqrt{ \frac{9}{4}}$

$= > x = \sqrt{ \frac{13+9}{4} } = \sqrt{ \frac{22}{4} } = \sqrt{ \frac{11}{2}}$

$hence \: x = + \sqrt{ \frac{11}{2} } \: and \\ \: \: \: \: \: \: \: \: \: \: \: \: \: x = - \sqrt{ \frac{11}{2} }$