Question

the quadratic equation ax square - 4ax plus 2a plus 1 = 0 has repeated root if a​

Answer 1

For equal roots, D = b^2 - 4ac = 0.

ax² - 4ax + (2a + 1) = 0

• a = a
• b = -4a
• c = 2a + 1

D = (-4a)² - 4(a)(2a + 1) = 0

=> 16a² - 8a² - 4a = 0

=> 8a² - 4a = 0

=> 4a(2a - 1) = 0

So either a = 0, or a = 1/2