Solve the following quadratic equation. √3 x2 + √2 x – 2√3 =0 I got this from the question bank released by Maharashtra board Please Answer About the author Caroline
Answer: [tex] \sqrt{3x}^{2} + \sqrt{2x} – 2 \sqrt{3} = 0[/tex] comparing [tex] \sqrt{3 {x}^{2} } + \sqrt{2x} – 2 \sqrt{3} = 0[/tex] with [tex]a {x}^{2} + bx + c = 0[/tex] [tex]a = \sqrt{3} \\ b = \sqrt{2} \\ c = – 2 \sqrt{3} [/tex] [tex] {b}^{2} – 4ac = ( \sqrt{2} )^{2} – 4 \times \sqrt{3} \times – 2 \sqrt{3} \\ = 2 + 6 \times 3 \\ = 2 + 18 \\ = 20[/tex] by formula method [tex] x = \frac{ – b + – \sqrt{ {b}^{2} – 4ac} }{2a} \\ x= \frac{- \sqrt{2} + – \sqrt{20} }{2 \sqrt{3} } \\ x = \frac{- \sqrt{2} + 2 \sqrt{5} }{2 \sqrt{3} } [/tex] [tex]or x = \frac{ – \sqrt{2} – 2 \sqrt{5} }{2 \sqrt{3} } [/tex] Reply
Answer:
[tex] \sqrt{3x}^{2} + \sqrt{2x} – 2 \sqrt{3} = 0[/tex]
comparing
[tex] \sqrt{3 {x}^{2} } + \sqrt{2x} – 2 \sqrt{3} = 0[/tex]
with
[tex]a {x}^{2} + bx + c = 0[/tex]
[tex]a = \sqrt{3} \\ b = \sqrt{2} \\ c = – 2 \sqrt{3} [/tex]
[tex] {b}^{2} – 4ac = ( \sqrt{2} )^{2} – 4 \times \sqrt{3} \times – 2 \sqrt{3} \\ = 2 + 6 \times 3 \\ = 2 + 18 \\ = 20[/tex]
by formula method
[tex] x = \frac{ – b + – \sqrt{ {b}^{2} – 4ac} }{2a} \\ x= \frac{- \sqrt{2} + – \sqrt{20} }{2 \sqrt{3} } \\ x = \frac{- \sqrt{2} + 2 \sqrt{5} }{2 \sqrt{3} } [/tex]
[tex]or x = \frac{ – \sqrt{2} – 2 \sqrt{5} }{2 \sqrt{3} } [/tex]