Question

Solve the following quadratic equations by completing the squares method.
2x² + x -4 = 0​

Answer 1

01

Step-by-step explanation:

2x2 +4(-7) =π9

/7:86

038) 9396+-_&#/₹+@

Answer 2

Step-by-step explanation:

$2 {x}^{2} + x - 4 = 0$

$⇒2 {x}^{2} + x = 4$

$⇒ {x}^{2} + \frac{1}{2} x = \frac{4}{2} = 2$

$⇒ {x}^{2} + 2. \frac{1}{2} . \frac{1}{2} x + {( \frac{1}{4} )}^{2} - {( \frac{1}{4} )}^{2} = 2$

$⇒(x + { \frac{1}{4}) }^{2} = 2 + \frac{1}{16} \: \\ = \frac{33}{16} = {( \frac{ \sqrt{33} }{4} )}^{2}$

$on \: taking \: square \: root$

$\: \: \: \: \: \: \: \: \: \: \: \: x + \frac{1}{4} = ±( \frac{ \sqrt{33} }{4} )$

$on \: taking \: ( + ) \: sign$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + \frac{1}{4} = \frac{ \sqrt{33} }{4}$

$⇒ \: \: \: \: \: \: \: \: x = \frac{ \sqrt{33} }{4} - \frac{1}{4} \\ \: \: \: \: \: \: \: \: \: \: = \frac{ \sqrt{33 } - 1}{4}$

$on \: taking \: ( - ) \: sign \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + \frac{1}{4} = - \frac{ \sqrt{33} }{4}$

$⇒x = \frac{ - \sqrt{33} }{4} - \frac{1}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - ( \sqrt{33} + 1) }{4}$

$∴ \: \: \: \: \: \: Roots \: of \: equation \: are$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ \sqrt{33} - 1 }{4}, \frac{ - 1 \sqrt{33} - 1 }{4} .$