How do you use the quadratic formula to solve #4x^2+16x+15=0#?

1 Answer
Nov 9, 2016

Answer:

See explanation...

Explanation:

#4x^2+16x+15 = 0#

is in the form:

#ax^2+bx+c = 0#

with #a=4#, #b=16# and #c=15#

The roots are given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-16+-sqrt(16^2-4(4)(15)))/(2*4)#

#color(white)(x) = (-16+-sqrt(256-240))/8#

#color(white)(x) = (-16+-sqrt(16))/8#

#color(white)(x) = (-16+-4)/8#

#color(white)(x) = (-4+-1)/2#

That is:

#x = -5/2# or #x = -3/2#

#color(white)()#
Footnote

In this particular example, I would probably have favoured an AC method rather than the quadratic formula.

Given:

#4x^2+16x+15 = 0#

Find a pair of factors of #AC=4*15=60# with sum #B=16#

The pair #10, 6# works.

Use this pair to split the middle term and factor by grouping:

#0 = 4x^2+16x+15#

#color(white)(0) = (4x^2+10x)+(6x+15)#

#color(white)(0) = 2x(2x+5)+3(2x+5)#

#color(white)(0) = (2x+3)(2x+5)#

Hence roots #x = -3/2# and #x = -5/2#

Note that the AC method is only applicable if the quadratic has rational zeros, whereas the quadratic formula will always give you the solution, even if the roots are Complex.