If equation $\displaystyle { 2x }^{ 2 }-2ax+4=0$ has 2 distinct roots, then
- $\displaystyle \left| a \right| > 2\sqrt { 2 } $
- $\displaystyle \left| a \right| < 2\sqrt { 2 } $
- $\displaystyle a=2\sqrt { 2 } $
-
None
Reveal answer
Fill a bubble to check yourself
A
Correct answer
Explanation
For distinct roots, the discriminant D > 0. D = (-2a)^2 - 4(2)(4) = 4a^2 - 32. 4a^2 - 32 > 0 implies a^2 > 8. Thus |a| > sqrt(8) = 2*sqrt(2).
AI explanation
For a quadratic equation to have two distinct real roots, the discriminant (b squared minus 4ac) must be greater than zero. Substituting the values from 2x squared minus 2ax plus 4 equals 0 gives (minus 2a) squared minus 4 times 2 times 4 is greater than 0. Simplifying this yields 4a squared minus 32 is greater than 0, so 4a squared is greater than 32, which means a squared is greater than 8. Taking the square root gives the absolute value of a is greater than 2 times the square root of 2.