How do you solve #23p = 5p² + 24#?

2 Answers
Jul 22, 2016

Answer:

# p = 3, p = 8/5#

Explanation:

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#

Jul 22, 2016

Answer:

#8/5 and 3#

Explanation:

#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#