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

2 Answers
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.

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