How do you solve #23p = 5p² + 24#?

Question

Isaiah Anthony · Accepted Answer

# p = 3, p = 8/5#

Solve quadratic equations by making them equal to 0.

#5p^2 -23p +24 =0#

Factorise: Find factors of 5 and 8 which add to give 23. Signs will be the same they are both negative.

Cross multiply the factors:

#"5 8"rArr 1 xx 8 = 8# #"1 3" rArr 5 xx3 = 15" "15+8 = 23#

#(5p-8)(p-3) = 0 " we have two factors"#

Either of the factors could be equal to 0.

#5p - 8 = 0," or "p-3 = 0# # 5p = 8" " p = 3# #p= 8/5#

Hazel Anglin · Answer

#8/5 and 3# #y = 5p^2 - 23p + 24.# Solve this quadratic equation by the new Transforming Method (Socratic Search). Method. First find the 2 real roots of the transformed equation, y' = p^2 - 23p + 120 = 0. Next, divide the answers by (a) to get the 2 real roots of y. Two real roots of y' have same sign (Rule of signs) Factor pairs of (ac = 120) --> (4, 30)(5, 24)(8, 15). This last sum is 23 = -b. The 2 real roots of y' are: 8 and 15. Back to original equation y, the 2 real roots are: #p1 = 8/a = 8/5 = 8/5#, and #p2 = 15/a = 15/5 = 3#

Genesis Allen · Answer

undefined To solve the equation 23p = 5p² + 24, follow these steps:

1. Rearrange the equation into a quadratic form: 5p² - 23p + 24 = 0.
2. Factor or use the quadratic formula to solve for p.

The quadratic formula is: $ p = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

In this equation:
a = 5,
b = -23,
c = 24.

Plug these values into the quadratic formula and solve for p.