Please help I don't understand!!

Question

in the equations listed, a, b, c, and d are all real numbers. Which of the equations would have solutions that are nonreal?

a. ax^2 = b

b. ax^2 + bx = 0

c. ax^2 + bx + c = 0

d. (ax + b)(cx + d) = 0

e. a(bx + c)^2 = d

Answer 1

a) solving for x we get x²=b/a, so x = ±√(b/a) will be NON-REAL if (b/a) is negative

b) ax² + bx = 0 Factor out an "x" and get x(ax+b) = 0

so x = 0 or ax+b=0, or x=-(b/a) both real numbers

c) ax² + bx + c = 0 This will have NON-REAL solutions when b²-4ac is negative

d) (ax + b)(cx + d) = 0

So ax+b = 0 or cx + d = 0 or

x = -(b/a) or x = -(d/c) both real numbers

e) a(bx+c)² = d solving for x: (bx+c)² = d/a

bx+c = ±√(d/a)

bx = ±√(d/a)-c

x = (1/b)[ ±√(d/a) - c ] this can also have NON-REAL solutions when (d/a) is negative.

Basically any solution can be NON-REAL if it involves the square root of a negative number.

Answer 2

You get a non-real number when you take the square root of a negative number.

(a)

ax² = b

x² = b/a.

Here, if b/a is negative, then the two solutions x = ±√(b/a) are imaginary.

(b)

ax² + bx = 0

x(ax + b) = 0

x = 0 or ax + b = 0.

Solving ax + b = 0 is routine and involves no square roots.

(d)

(ax + b)(cx + d) = 0

ax + b = 0 or cx + d = 0.

Same thing here.

a(bx + c)² = d

(bx + c)² = d/a

If d/a is negative then when you take square roots, you encounter the fact that bx + c is imaginary. When you add –c to that imaginary number, you get a complex number in which the real part is not 0, so it's not a "pure" imaginary number.

For part (c), remember the standard formula for solving quadratic equations:

If ax² + bc + c = 0 and a ≠ 0, then x = (–b ± √(b² – 4ac) ) / 2.

If the thing whose square root you're taking, which is b² – 4ac, is negative, then the solutions are not real.