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solve for x by factorisation method. 3x²+√21x-14=0

Posted by Syeda Sana 7 years, 2 months ago

CBSE > Class 10 > Mathematics

3 answers

Naveen Sharma 7 years, 2 months ago

Ans. $3 x^{2} + \sqrt{21} x - 14 = 0$

=> $3 x^{2} + 2 \sqrt{21} x - \sqrt{21} x - 14 = 0$

=> $\sqrt{3} x (\sqrt{3} x - 2 \sqrt{7}) - \sqrt{7} (\sqrt{3} x - 2 \sqrt{7)} = 0$

=> $(\sqrt{3} x - \sqrt{7}) (\sqrt{3} x - 2 \sqrt{7}) = 0$

$E i t h e r \sqrt{3} x - \sqrt{7} = 0 o r \sqrt{3} x - 2 \sqrt{7} = 0$

$x = \frac{\sqrt{7}}{\sqrt{3}} o r \frac{2 \sqrt{7}}{\sqrt{3}}$

0Thank You

Rashmi Bajpayee 7 years, 2 months ago

$3 x^{2} + \sqrt{21} x - 14 = 0$

$3 x^{2} + 2 \sqrt{21} x - \sqrt{21} x - 14 = 0$

$\sqrt{3} x (\sqrt{3} x + 2 \sqrt{7}) - \sqrt{7} (\sqrt{3} x + 2 \sqrt{7}) = 0$

$(\sqrt{3} x + 2 \sqrt{7}) (\sqrt{3} x - \sqrt{7}) = 0$

$\sqrt{3} x + 2 \sqrt{7} = 0 a n d \sqrt{3} x - \sqrt{7} = 0$

$x = - 2 \sqrt{\frac{7}{3}} a n d x = \sqrt{\frac{7}{3}}$

0Thank You

Rashmi Bajpayee 7 years, 2 months ago

3x² + 2 $\sqrt{21}$ x - $\sqrt{21}$ x - 14 = 0

$\sqrt{3} x (\sqrt{3} x + 2 \sqrt{7}) - \sqrt{7} (\sqrt{3} x + 2 \sqrt{7}) (\sqrt{3} x + 2 \sqrt{7}) (\sqrt{3} x - \sqrt{7}) \sqrt{3} x + 2 \sqrt{7} = 0, \sqrt{3} x - \sqrt{7} = 0 x = \frac{- 2 \sqrt{7}}{\sqrt{3}} = - 2 \sqrt{\frac{7}{3}}, x = \frac{\sqrt{7}}{\sqrt{3}} = \sqrt{\frac{7}{3}}$