Question

solve:- 3x^2 = 2x -8
from quadratic equation​

Answer 1

$\huge\bold\red{ANSWER:-}$

$3x {}^{2} = 2x - 8 \\ .°. 3x {}^{2} - 2x + 8 = 0 \\ using \: \: quadratic \: equation \\ x = \frac{ - ( - 2)± \sqrt{( - 2) {}^{2} - 4 \times 3 \times 8 } }{2 \times 3} \\ x = \frac{2± \sqrt{4 - 96} }{6} \\ x = \frac{2± \sqrt{ - 92} }{6}$

THE SQUARE ROOT OF A NEGATIVE NUMBER DOES NOT EXIST IN THE SET OF REAL NUMBERS.

.°.X € R.

$\huge\bold\red{THANKS.}$

Answer 2

Answer:

$\large\boxed{\sf{x = \frac{1 + \sqrt{23} \iota}{3} }}$

Step-by-step explanation:

It's given a quadratic equation,

$\sf{3 {x}^{2} = 2x - 8 }$

Moving all the terms on LHS,

$\sf{= > 3 {x}^{2} - 2x + 8 = 0} \\ \\ \sf{= > \frac{3 {x}^{2} - 2x + 8 }{3} = \frac{0}{3} } \\ \\ \sf{ = > {x}^{2} - 2 \times x \times \frac{1}{3} + \frac{8}{3} = 0 }\\ \\ \sf{= > {x}^{2} - 2 \times x \times \frac{1}{3} + {( \frac{1}{3} )}^{2} + \frac{8}{3} - {( \frac{1}{3} )}^{2} = 0 }\\ \\ \sf{ = > {(x - \frac{1}{3} )}^{2} + \frac{8 \times 3}{3 \times 3} - \frac{1}{9} = 0 }\\ \\ \sf{ = > {(x - \frac{1}{3}) }^{2} + \frac{24 - 1}{9} = 0} \\ \\ \sf{= > {(x - \frac{1}{3} )}^{2} + \frac{23}{9} = 0} \\ \\ \sf{ = > {(x - \frac{1}{ 3}) }^{2} = - \frac{23}{9} } \\ \\ \sf{ = > x - \frac{1}{3} = \sqrt{ - \frac{23}{9} }} \\ \\ \sf{ = > x - \frac{1}{3} = \sqrt{ \frac{23}{9} } \times \sqrt{ - 1} } \\ \\ \sf{= > x - \frac{1}{3} = \frac{ \sqrt{23} \iota}{3}} \\ \\ \sf{ = > x = \frac{1}{3} + \frac{ \sqrt{23} \iota}{3}} \\ \\ \sf{ = > x = \frac{1 + \sqrt{23} \iota}{3} \: \: \: \: \: ( \iota= iota)}$

Hence, the value of x is complex number.