How do you solve #(4x-2)/(x-6) = -x/(x+5)# and find any extraneous solutions?

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Answer 1 · 2016-07-25T14:23:41

The solns. are #x_1~=0.655, and, x_2~=-3.055#.

#(4x-2)/(x-6)=-x/(x+5)#

By Cross-Multiplication, #(4x-2)(x+5)=-x(x-6)#

By Expansion, # 4x^2+20x-2x-10=-x^2+6x#

By Transposition, #4x^2+18x-10+x^2-6x=0#

#:. 5x^2+12x-10=0.............(1)#

We use the Formula to find the roots of this Qudr. Eqn. :-

The roots #alpha, beta# of the Gen. Qudr. Eqn. #ax^2+bx+c=0# are

#alpha=(-b+sqrt(b^2-4ac))/(2a), beta=(-b-sqrt(b^2-4ac))/(2a)#

In our case, since, #a=5, b=12, c=-10#, we get,

#alpha=(-12+sqrt(12^2-4*5*(-10)))/(2*5)=(-12+sqrt(144+200))/10#

#:. alpha=(-12+sqrt344)/10, and, beta=(-12-sqrt344)/10#

Taking, #sqrt344~=18.55, alpha~=0.655, and, beta~=-3.055#

Thus, the solns. are #x_1=alpha~=0.655, and, x_2=beta~=-3.055#.