Question

find the zeroes of 4√3x² + 5x​

Answer 1

$4 \sqrt{3x² }+ 5x = 0 \\ = > 4 \sqrt{3x² } = - 5x \\ = > \sqrt{3x² } = \frac{ - 5x}{4} \\ = > 3 {x}^{2} = { (\frac{ - 5x}{4} })^{2} \\ = >3 {x}^{2} = \frac{25 {x}^{2} }{16} \\ = > 3 {x}^{2} \times 16 = 25 {x}^{2} \\ = > 48 {x}^{2} = 25 {x}^{2} \\ = > 48 {x}^{2} - 25 {x}^{2} = 0 \\ = > 23 {x}^{2} = 0 \\ = > {x}^{2} = - 23 \\ = > x = \sqrt{ - 23} \\ = > x = - 4.79583152$

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