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# In the Following, Determine Whether the Given Quadratic Equation Have Real Roots and If So, Find the Roots: 16x2 = 24x + 1 - Mathematics

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#### Question

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

16x2 = 24x + 1

#### Solution

We have been given,

16x2 = 24x + 1

16x2 - 24x - 1 = 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 = 16, b = -24 and c = -1.

Therefore, the discriminant is given as,

D = (-24)2 - 4(16)(-1)

= 576 + 64

= 640

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=(-(-24)+-sqrt640)/(2(16))

=(24+-8sqrt10)/32

=(3+-sqrt10)/4

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

x=(3+sqrt10)/4

Also,

x=(3-sqrt10)/4

Therefore, the roots of the equation are (3+sqrt10)/4and (3-sqrt10)/4

Is there an error in this question or solution?