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