Question

find the roots of the following quadratic equation given in question 1 above by applying the. a)2xsquare-7x+3=0 b)2xsquare +x-4=0 c)4xsqrare+4√3x+3=0 d)2xsquare+x+4=0​

Answer 1

Answer:

plz have a look hope it may help you

Answer 2

Step-by-step explanation:

$2 {x}^{2} - 7x + 3 = 0 \\ 2 {x}^{2} - 6x - x + 3 = 0 \\ 2x(x - 3) - 1(x - 3) = 0 \\ (x - 3)(2x - 1) = 0 \\ x - 3 = 0 \: \: or \: \: 2x - 1 = 0 \\ x = 3 \: \: or \: \: 2x = 1 \\ \boxed{x = {\underline{ \underline{3}}} \: \: or \: \: x = {\underline{ \underline \frac{1}{2} }}} \\ \\ 2 {x}^{2} + x - 4 = 0 \\ x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - ( + 1)± \sqrt{ {1}^{2} - 4(2)( - 4) } }{2 \times 2} \\ x = \frac{ - 1± \sqrt{1 - ( - 32)} }{4} \\ x = \frac{ - 1± \sqrt{1 + 32} }{4} \\ x = \frac{ - 1± \sqrt{33} }{4} \\ \boxed{x = { \underline{ \underline{ \frac{ - 1 + \sqrt{33} }{4} }}} \: \: or \: \: x = { \underline{ \underline{ \frac{ - 1 - \sqrt{33} }{4} }}}}$