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# In the Following, Determine Whether the Given Quadratic Equation Have Real Roots and If So, Find the Roots: 3a2x2 + 8abx + 4b2 = 0, A ≠ 0 - Mathematics

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#### Question

In the following, determine whether the given quadratic equation have real roots and if so, find the roots:

3a2x2 + 8abx + 4b2 = 0, a ≠ 0

#### Solution

We have been given, 3a2x2 + 8abx + 4b2 = 0

Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:

D = b2 - 4ac

Now, according to the equation given to us, we have,a = 3a2, b = 8ab and c = 4b2.

Therefore, the discriminant is given as,

D = (8ab)2 - 4(3a2)(4b2)

= 64a2b2 - 48a2b2

= 16a2b2

Since, in order for a quadratic equation to have real roots, D ≥ 0.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

x=(-b+-sqrtD)/(2a)

Therefore, the roots of the equation are given as follows,

x=(-(8ab)+-sqrt16a2b2)/(2(3a2))

=(-8ab+-4ab)/(6a^2)

=(-4b+-2b)/(3a)

Now we solve both cases for the two values of x. So, we have,

x=(-4b+2b)/(3a)

=-(2b)/(3a)

Also,

x=(-4b-2b)/(3a)

=(-2b)/a

Therefore, the roots of the equation are -(2b)/(3a) and (-2b)/a.

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