Question

find zeroes for the quadratic polynomial 2√3x²-5x+3

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x= \frac{5 \pm4.07 \iota}{4 \sqrt{3}}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies 2 \sqrt{3}{x}^{2} - 5x + 3 = 0 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Value \: of \: x = ?$

• According to given question :

$\tt: \implies 2 \sqrt{3} x^{2} - 5x + 3 = 0 \\ \\ \tt\circ \: a = 2 \sqrt{3} \\ \\ \tt\circ \: b = - 5 \\ \\ \tt \circ \:c = 3 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies D= {b}^{2} - 4ac\\ \\ \tt: \implies D = {( - 5)}^{2} - 4 \times 2 \sqrt{3} \times 3 \\ \\ \tt: \implies D = 25 - 24 \sqrt{3} \\ \\ \tt: \implies D= 25 - 41.6 \\ \\ \tt: \implies D = - 16.6 \\ \\ \tt: \implies x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt: \implies x = \frac{ - ( - 5) \pm \sqrt{ - 16.6} }{2 \times 2 \sqrt{3} } \\ \\ \tt: \implies x = \frac{5 \pm \sqrt{16.6} \iota }{4 \sqrt{3} } \\ \\ \green{\tt: \implies x= \frac{5 \pm4.07 \iota}{4 \sqrt{3} } }$

Answer 2

Ans:

Solve by splitting middle term

=> 2√3x² - 5x - √3 = 0

=> 2√3x² - 6x + x - √3 = 0

=> 2√3x (x - √3) + 1 (x - √3) = 0

=> (x - √3) (2√3x + 1) = 0

Either

=> x - √3 = 0

=> x = √3

or

=> 2√3x + 1 = 0

=> 2√3x = -1

=> x = -1/2√3

√3 and -1/2√3 are the zeros