Question

find the roots of quadratic equation x square+root2x-12=0​

Answer 1

Answer:

Hope this helps!

x² + √2 x - 12 = 0

=> x = ( -√2 ± √( √2² - 4×1×(-12) ) ) / 2

     = ( -√2 ± √( 2 + 48 ) ) / 2

     = ( -√2 ± √50 ) / 2

     = ( -√2 ± 5√2 ) / 2

     = -6√2 / 2   or 4√2 / 2

     = -3√2  or 2√2

Answer 2

root of quadratic equation

$a {x}^{2} + bx + c = 0$

is

$x = \frac{ - b \: \binom{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a}$

so in given equation

x^2 + (root 2 )x -12 =0

a = 1

b = root 2

c = -12

so b^2-4ac = (root 2)^2 - 4 ×1×(-12)= 2 + 48 = 50

so now roots are

$x = \frac{ - \sqrt{2} \: \binom{ + }{ - } \sqrt{ 50}}{2} \:$

$x = \frac{ - \sqrt{2} \: + 5\sqrt{ 2}}{2} \:$

and

$x = \frac{ - \sqrt{2} \: - 5\sqrt{ 2}}{2} \:$