Question

Check wheather x=-root 2,-2root2 are roots of x2+root 2x -4=0

Answer 1

Hey friend,

Here is the solution :-

$given : x = ( - \sqrt{2} ) \: and \: ( - 2 \sqrt{2} ) \\ \\ eqn. = {x}^{2} + \sqrt{2} x - 4 = 0 \\ \\ when \: x = ( - \sqrt{2} ) : - \\ \\ {( - \sqrt{2} })^{2} + \sqrt{2} \times ( - \sqrt{2} ) - 4 = 0 \\ 2 + 2 - 4 = 0 \\ 4 - 4 = 0 \\ \\ when \: x = ( - 2 \sqrt{2} ) \\ \\ ( - 2 \sqrt{2} )^{2} + \sqrt{2} \times ( - 2 \sqrt{2} ) - 4 = 0 \\ 8 + ( - 4) - 4 = 0 \\ 8 - 4 - 4 = 0 \\ 8 - 8 = 0$

Hence, both are the roots of the equation.

✨I hope this will help you.... ✨

Answer 2

hi mate,

solution:

given: x = √2

The given equation is :

→x² + √2 x - 4 =0

By splitting the middle term,we can write the equation as

→ x² + 2 √2 x - √2 x - 4=0

→ x² + 2 √2 x - √2 x - 2√2×√2 =0

→ x ( x + 2 √2 ) - √2 (x + 2√2) = 0

→ (x - √2) (x + 2 √2) = 0

→ x - √2 =0 and x + 2√2=0

→ x= √2 and x = - 2 √2 , are solution of the equation.

i hope it helps you.