Question

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

16x² = 24x + 1

Answer 1

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)/4`and `(3-sqrt10)/4`