Question

quardic equations class ten icse
√3x×x +10x-8√3=0​

Answer 1

Answer:

-7/root 3 and 2/ root 3

Step-by-step explanation:

the value of b^2-4ac =(10)^2- 4×root 3 ×-8 root 3

=100+96=196

196>0 so we have roots

using -b plus or minus root b^2-4ac/2a

one root is -b+ root b^2-4ac/2a = -10+ root 196 /2 root 3

=4/2 root 3 = 2/root 3

other root is -b- root b^2-4ac /2a =-10- root 196 /2 root 3

=-14/ 2 root 3= -7/ root 3