Question

Solve the equation by complete square method 2x^2+10x+3=0

Answer 1

Answer:

Step-by-step explanation:

2x² + 10x + 3=0

x²+5x+3/2=0. (Dividing by 2)

x²+5x=-3/2

x²+5x+(5/2)²=-3/2+(5/2)². (The coefficient of X should be divided by 2 and then it's square should be added to both sides)

(X +(5/2)²)=(-3/2) + 24/4=19/4 (here in the LHS we will take the x² and the (5/2)² together)

X+5/2=+-√19/2

X= √19-5/2 or -√19-5/2

Answer 2

Given,

2x^2 + 10x + 3 = 0

$\underline \bold{solution}$


$\frac{ {2x}^{2} }{2} + \frac{10x}{2} + \frac{3}{2} = 0 \\ \\ = > {x}^{2} + 5x = - \frac{3}{2} \\ \\ = > {x}^{2} + 5x + {(5 \times \frac{1}{2} )}^{2} = \frac{ - 3}{2} + {(5 \times \frac{1}{2} )}^{2} \\ \\ = > {x}^{2} + 5x + ({ \frac{5}{2} )}^{2} = \frac{ - 3}{2} + \frac{25}{4} \\ \\ = > {(x + \frac{5}{2} )}^{2} = \frac{ - 6 + 25}{4} \\ \\ = > {(x + \frac{5}{2} )}^{2} = \frac{19}{4} \\ \\ = > x + \frac{5}{2} = \sqrt{ \frac{19}{4} } \\ \\ = > x + \frac{5}{2} = \frac{ \sqrt{19} }{2} \\ \\ = > x = ± \frac{ \sqrt{19} }{2} - \frac{5}{2} \\ \\ \\ \\ \\ \\ x = \frac{ \sqrt{19} }{2} - \frac{5}{2} \\ \\ = \frac{ \sqrt{19} - 5}{2} \\ \\ \\ \\ x = \frac{ - \sqrt{19} }{2} - \frac{5}{2} \\ \\ = \frac{ - \sqrt{19} - 5 }{2}$



$\huge \boxed{answer = \frac{ \sqrt{19} + 5}{2} \: and \frac{ - \sqrt{19} - 5}{2} }$