Question

Slove 2x² +x+ 4=0
formula method​

Answer 1

Answer:

_____________________________________________________________

Given & To find:

Solve for x using completing the square method,.

2x² + x - 4 = 0,

_____________________________________________________________

As, we know that,

(a + b)² = a² + 2ab + b²

_____________________

We are intended to find the value of x, using this formula,.

=> For doing that, we need to bring this equation to that form (a² + 2ab + b² for some value a & b.,.)

We take an attempt to bring this equation to that form,

=> 2x² + x - 4 = 0,

=> (Dividing both the sides by ),

=>

 (Multiplying & dividing the  middle term by 2 to get the term 2ab (here a = x, ))

So, we have the form,

a² + 2ab + c  (for some value c, here it is - 4), we need b² to bring this equation to that form,

By adding both the sides ,

we get,

=>

It can be, reduced to the form (a + b)²

=>

=> 

=>

=>

By moving the square from LHS to RHS, we get,

=>

 

=> (or)

=> (or)

=> (or)

It's completed,.

____________________________________________________________

                                 Hope it Helps !!

=> Mark as Brainliest,..

Answer 2

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

$\bold\blue {x = -8\:or\:7.5}$

$\bold\red{\underline{\underline {Step-by-step \:Explaination :}}}$

$2 {x}^{2} + x + 4 = 0 \\ \\ here \: a = 2 \\ \: \: \: \: \: \: \: \: \: \: b = 1 \\ \: \: \: \: \: \: \: \: \: c = 4 \\ \\ x = \frac{ - b∓( {b}^{2} - 4ac )}{2a} \\ \: \: \: = \frac{ - 1∓( {1}^{2} - 4 \times 2 \times 4) }{2 \times 2} \\ \: \: \: = \frac{ - 1∓(1 - 32)}{4} \\ \: \: \: = \frac{ - 1∓( - 31)}{4} \\ \\ x1 = \frac{ - 1 + ( - 31)}{4} \\ \: \: \: \: \: \: = \frac{ - 1 - 31}{4} \\ \: \: \: \: \: \: = \frac{ - 32}{4} \\ \: \: \: \: \: \: = - 8 \\ \\ x2 = \frac{ - 1 - ( - 31)}{4} \\ \: \: \: \: \: \: = \frac{ - 1 + 31}{4} \\ \: \: \: \: \: \: = \frac{30}{4} \\ \: \: \: \: \: \: = 7.5$

$\bold\green{\boxed{Hence,\: The\:Value\:of\:x\:is\:-8\:or\:7.5}}$

$<marquee>  ❤♥ Please follow me and mark as brainliest. ♥❤   </marquee>$