If a = 0.125 then what is value of 4a2−4a+1+3a\sqrt{4a^2-4a+1}+3a4a2−4a+1+3a ? A1.500B1.125C1.250D1.225Answer: B. 1.125 Read Explanation: Solution:Given:a = 0.125Concept used:(A - B)2 = A2 - 2AB + B2Calculation:Now,4a2−4a+1\sqrt{4a^2-4a+1}4a2−4a+1⇒(1−4a+4a2)⇒\sqrt{(1-4a+4a^2)}⇒(1−4a+4a2)⇒(1−2a)2⇒\sqrt{(1-2a)^2}⇒(1−2a)2⇒(1−2a)⇒(1-2a)⇒(1−2a)So,4a2−4a+1+3a\sqrt{4a^2-4a+1}+3a4a2−4a+1+3a⇒ (1 - 2a) + 3a⇒ 1 + a⇒ 1.125∴ The required value of 4a2−4a+1+3a\sqrt{4a^2-4a+1}+3a4a2−4a+1+3a is 1.125. Read more in App