How do you solve #4x^2+17x+15=0#?

3 Answers

Answer:

#x = -3# or #x = -5/4#

Explanation:

#4x^2 + 17x+15=0#

Do

#4x^2+12x+5x+15=0#

Then

#4x(x+3)+5(x+3)=0#

#(4x+5)(x+3)=0#

So

#x =-3" "# or #" "x=-5/4#

Jun 11, 2018

Answer:

#-3, and -5/4#

Explanation:

Method 1. Use the improved quadratic formula
#y = 4x^2 + 17x + 15 = 0#.
#D = d^2 = b^2 - 4ac = 289 - 240 = 49# --> #d = +- 7#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = - 17/8 +- 7/8#
#x1 = -24/8 = - 3#
#x2 = -10/8 = -5/4#
Method 2. Use the new Transforming Method (Socratic, Google Search)
Transformed equation: #y' = x^2 + 17x + 60#.
Proceeding. Find the 2 real roots of y', then, divide them by a = 4.
Find 2 real roots, bot negative (Rule of signs), knowing the sum (-b = -17), and the product (ac = 60). They are - 5 and - 12.
The 2 real roots of y are:
#x1 = -5/a = -5/4#, and #x2 = -12/a = -12/4 = -3#

Jun 11, 2018

Answer:

#x-3 or x = -5/4#

Explanation:

Factorise the quadratic trinomial.

Find factors of #4 and 15# whose products add to #17#

#" "4 and 15#
#" "darr" "darr#
#" "4" "5" "rarr 1 xx 5 =5#
#" "1" "3" "rarr 4xx3 = ul12#
#color(white)(xxxxxxxxxxxxxxxxx)17#

The signs are all positive.

#(4x+5)(x+3)=0#

Either factor can be equal to #0#

If #4x+5=0" "rarr x =-5/4#

If #x+3=0" "rarr x =-3#