How do you solve #y^2-90=13y#?

2 Answers
Jul 6, 2017

Answer:

I used the quadratic formula, and found:

#y=18 or y=-5#

Explanation:

We have:

#y^2-90=13y#

Rearrange into standard form:

#y^2-13y-90=0#

Standard form has the powers of the pronumeral in decreasing order on the left and #0# on the right. It can be written as:

#ay^2+by+c=0#, where #a#, #b# and #c# are the coefficients of the powers of the pronumeral.

In this case, #a=1#, #b=-13# and #c=-90#.

We often use #x# as the pronumeral, but we treat #y# just the same way. The quadratic formula in terms of #y# is just:

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

Substituting in the values from the equation:

#y = (13+-sqrt((-13)^2-4(1)(-90)))/(2(1))#

#y = (13+-sqrt(169+360))/2 = (13+-23)/2#

Therefore:

#y=18 or y=-5#

This quadratic could also be solved by factorising.

Jul 6, 2017

Answer:

#y=18 and -5#

Explanation:

Write as:#" "y^2-13y-90=0#

This is no different to a quadratic in #x# in format. The only difference is that it is in #y#. So the whole thing is:

#x=0=y^2-13y-90#

A quadratic in #x# where #ax^2>0# gives #uu#
A quadratic in #y# where #ay^2>0# gives #sub#
So it is rotating #uucolor(white)(.)90^o# clockwise.

Lets check the whole number factors of 90 to see if any of them give a difference of 13

#1xx90=90#
#2xx45=90#
#3xx30=90#
#4xx?=90larr" ? will not be a whole number factor" #
#color(purple)(5xx18=90 larr" difference is 13")#

The 90 is negative so the two numbers are of opposite sign.
The 13 is negative so the greater of the two is negative giving:

#(y-18)(y+5)=0#

Condition 1 : #" "y-18=0 " "=>" "y=+18#
Condition 2: #" "y+color(white)(1)5=0" "=>" "y=-color(white)(1)5#

Thus #y=18 and -5#

Tony B