How do you solve #15r ^ { 2} - 21r - 18= 0#?

2 Answers
Shwetank Mauria
Apr 11, 2017

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

Explanation:

All the monomials in the equation #15r^2-21r-18=0# are divisible by #3# and hence dividing both sides by #3#, we get

#5r^2-7r-6=0#

To solve this, we should first factorize LHS, where we have a quadratic polynomial. As the coefficients of #r^2# and constant term are opposite in sign and their product is #5xx6-30#, we should identify two factors of #30#, whose difference is #7#, the coefficient of middle term.

It is apparent that these are #10# and #3# and we should split middle term accordingly. Hence, this becomes

#5r^2-10r+3r-6=0#

or #5r(r-2)+3(r-2)=0#

or #(5r+3)(r-2)=0#

#:.# either #5r+3=0# i.e. #5r=-3# and #r=-3/5#

or #r-2=0# i.e. #r=2#

Note : In case signs of the coefficients of #r^2# and constant term are same , you identify two factors of their product, whose sum is the coefficient of middle term.

AA_123
Apr 11, 2017

#r=2# and #r=-3/5#

Explanation:

First of all divide everything by #3# to simplify the equation

#(15r^2-21r-18)/3=0/3#

#5r^2-7r-6=0#

Use the quadratic formula: #r=(-bpmsqrt(b^2-4ac))/(2a)#

In this case #b=-7, a=5, c=-6#

#r=(-(-7)pmsqrt((-7)^2-4(5)(-6)))/(2(5))#

#color(white)r=(7pmsqrt(49+120))/10#

#color(white)r=(7pmsqrt(169))/10#

#color(white)r=(7pm13)/10#

#r=(7+13)/10# or #(7-13)/10#

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

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