Question

The equation 6x2 - 132 +5 -0 has solutions of the form

Answer 1

Answer:

(A)

$N = -b = -(-13) = 13$

$D =b^2 -4ac = (-13)^2 - 4(6)(5) = 169- 120 = 49$

$M = 2a = 2(6) = 12$

(B)

$x = (\frac{5}{3} , \: \frac{1}{2})$

Step-by-step explanation:

The given equation is

$6x^2 - 13x + 5 = 0$

The solution is of the form as given by

$x=\frac{N\pm\sqrt{D}}{M}$

(A) Use the quadratic formula to solve this equation and find the appropriate integer values of N, M and D. Do not worry about simplifying the VD yet in this part of the problem.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equations of N, M and D are

$N = -b$

$D =b^2 -4ac$

$M = 2a$

The values of a, b and c are

$a = 6 \\\\b = -13 \\\\c = 5$

So,

$N = -b = -(-13) = 13$

$D =b^2 -4ac = (-13)^2 - 4(6)(5) = 169- 120 = 49$

$M = 2a = 2(6) = 12$

(B) Now simplify the radical and the resulting solutions. Enter your answers as a list of integers or reduced fractions, separated with commas. Example: -5/2-3/4

N = 13

D = 49

M = 12

$x=\frac{13\pm\sqrt{49}}{12}$

$x=\frac{13\pm7}{12}$

$x=\frac{13+7}{12}$  and $x=\frac{13-7}{12}$

$x=\frac{20}{12}$  and $x=\frac{6}{12}$

$x=\frac{5}{3}$  and $x=\frac{1}{2}$

$x = (\frac{5}{3} , \: \frac{1}{2})$