Q15 of 38 Page 190

Let’s find the roots of the equations (i.e. let’s solve the equations).

3(x-4)2+ 5(x-3)2 = (2x-5) (4x-1)-40


3(x-4)2+ 5(x-3)2 = (2x-5) (4x-1)-40


⇒ 3(x2-8x+16)+5(x2-6x+9)=2x(4x-1)-5(4x-1)-40


⇒ 3x2-24x+48+5x2-30x+45=8x2-2x-20x+5-40


⇒ 8x2-54x+93=8x2-22x-35


⇒ 93+35=54x-22x


⇒ 32x=128



=4


This is the required root of the given equation.


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