. If a, b, c are real numbers such that ac # 0 then show that at least one of the equations
ax + bx+ c=0 and -ax + bx+ c=0 has real
roots.

  if one equation has no real roots, the other must have real roots. Hence, at least one equation has real roots.

Step-by-step explanation:

The discriminant of the first equation is ...

  d1 = b^2 -4ac

The discriminant of the second equation is ...

  d2 = b^2 +4ac

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Suppose the first equation has no real roots. Then ...

  d1 < 0

  b^2 -4ac < 0

  b^2 < 4ac

We know that b^2 is non-negative, so this means 4ac is positive. For that case,

  d2 = b^2 +4ac

is the sum of a positive number and a non-negative number so will be positive. When the discriminant is positive, there are two real roots.

When the first equation has no real roots, the second one must have two real roots.

__

Suppose the second equation has no real roots. Then ...

  d2 < 0

  b^2 +4ac < 0

  b^2 < -4ac

Again, b^2 is non-negative, so -4ac must be positive. For this case, the sum ...

  d1 = b^2 -4ac

is the sum of a non-negative and a positive number, so will be positive. The positive discriminant means there are two real roots.

When the second equation has no real roots, the first one must have two real roots.

One of these equations will have real roots, or not. If not, the other must have (distinct) real roots.

answer: the anser is 3

step-by-step explanation:

c- 2o is equal to 9