If #2((3x+ 1)/(2x- 5))^2 + (3x + 1)/(2x- 5) - 6= 0#, then what is the value of #x#?
2 Answers
Explanation:
Let
#2u^2 + u - 6 = 0#
#2u^2 + 4u - 3u - 6 = 0#
#2u(u + 2) - 3(u + 2) = 0#
#(2u - 3)(u + 2) =0#
#u = 3/2 or -2#
We now return to our original variable
Case 1:
#3/2 = (3x + 1)/(2x - 5)#
#3(2x - 5) = 2(3x + 1)#
#6x - 15 = 6x + 2#
#0x = 17#
#x = O/#
Case 2:
#-2 = (3x + 1)/(2x - 5)#
#-2(2x - 5) = 3x + 1#
#-4x + 10 = 3x + 1#
#9 = 7x#
#x = 9/7#
Hopefully this helps!
Explanation:
For starters, you know that
In other words, you need to have
#2x - 5 != 0#
#x != 5/2#
Now, let's say that
#(3x+ 1)/(2x - 5) = y" "color(darkorange)("(*)")#
The original equation becomes
#2y^2 + y - 6 = 0#
You can solve this quadratic equation by using the quadratic formula, which for a general-form quadratic equation
#color(blue)(ul(color(black)(ax^2 + bx + c = 0)))#
looks like this
#color(blue)(ul(color(black)(x_(1,2) = (-b +- sqrt(b^2 - 4 * a * c))/(2 * a))))#
In your case, you have
#{(a = 2), (b = 1), (c = - 6) :}#
and so
#y_(1,2) = (-1 +- sqrt( 1^2 - 4 * 2 * (-6)))/(2 * 2)#
#y_(1,2) = (-1 +- sqrt(1 + 48))/4#
#y_(1,2) = (-1 +- sqrt(49))/4 implies {( y_1 = (-1 - 7)/4 = -2), (y_2 = (-1 + 7)/4 = 3/2) :}#
Take both values of
#ul("For" color(white)(.)y = -2)#
#(3x + 1)/(2x - 5) = -2# This is equivalent to
#3x + 1 = -2 * (2x - 5)#
#3x + 1 = - 4x + 10# which gets you
#7x = 9 implies x = 9/7#
#ul("For" color(white)(.)y = 3/2)#
#(3x + 1)/(2x - 5) = 3/2# This is equivalent to
#3x + 1 = 3/2 * (2x - 5)#
#color(red)(cancel(color(black)(3x))) + 1 = color(red)(cancel(color(black)(3x))) - 15/2#
#1 != -15/2#
Therefore, you can say that the original equation has
Do a quick double-check to make sure that the calculations are correct
#2 * ((3 * 9/7 + 1)/(2 * 9/7 - 5))^2 + (3 * 9/7 + 1)/(2 * 9/7 - 5) - 6 = 0#
#2 * (((27 + 7)/color(red)(cancel(color(black)(7))) )/( (18 - 35)/color(red)(cancel(color(black)(7)))))^2 + ((27 + 7)/color(red)(cancel(color(black)(7))) )/( (18 - 35)/color(red)(cancel(color(black)(7)))) - 6 = 0#
#2 * (34/(-17))^2 + (34/(-17)) - 6 = 0#
#2 * (-2)^2 + (- 2) - 6 = 0#
#2 * 4 - 8 = 0 " " color(darkgreen)(sqrt())#