Question

The equation 6x2 - 132 +5 -0 has solutions of the form
NVD
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.
N = ]:D
M
(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
Preview

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 $x=\frac{1}{2}$

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