The zeros of thepolynomial x^2-√2x-121)√2,-√22)3√2,-2√23) -3√2,2√24) 3√2,2√2 About the author Katherine
Answer: option B is correct Step-by-step explanation: We have, f(x) = x2 ‒ √2x – 12 Now, put f(x) = 0 x2 ‒ √2x – 12 = 0 x2 – 3√2x + 2√2x – 12 = 0 x(x ‒ 3√2) + 2√2(x – 3√2)=0 (x ‒ 3√2) + (x + 2√2) = 0 Thus, x= 3√2 , -2√2 Reply
Answer:
option B is correct
Step-by-step explanation:
We have, f(x) = x2 ‒ √2x – 12
Now, put f(x) = 0
x2 ‒ √2x – 12 = 0
x2 – 3√2x + 2√2x – 12 = 0
x(x ‒ 3√2) + 2√2(x – 3√2)=0
(x ‒ 3√2) + (x + 2√2) = 0
Thus, x= 3√2 , -2√2