Question

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

Answer 1

`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.

Answer 2

`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`