Question

write the equation
$(2x - 3)(6x - 1)(3 x - 2)(x - 2) - 5 = 0$
as a quadratic equation in another variable y​

Answer 1

Step-by-step explanation:

(2x - 3)(6x - 1)(3 x - 2)(x - 2) - 5 = 0solve the equation by partly factored polynomials

(2x-3)(6x-1)(3x-2)(x-2)-5=0

Rearranging we get

(2x-3)(3x-2)(6x-1)(x-2)-5=0

(6x^2-13x+6)(6x^2-3x+2)-5=0

Take 6x^2-13x=y

\implies\:(y+6)(y+12)-7=0

\implies\:(y^2+18y+72)-7=0

\implies\:y^2+18y+65=0

\implies\:(y+13)(y+5)=0

\implies\:y=-5,-13

case(i): y=-5

6x^2-13x=-5

6x^2-13x+5=0

(3x-5)(2x-1)=0

\implies\:x=\frac{1}{2},\frac{5}{3}

case(ii): y=-13

6x^2-13x=-13

6x^2-13x+13=0

x=\frac{13+\sqrt{169-312}}{12},\frac{13-\sqrt{169-312}}{12}

x=\frac{13+\sqrt{-143}}{12},\frac{13-\sqrt{-143}}{12}

x=\frac{13+i\sqrt{143}}{12},\frac{13-i\sqrt{143}}{12}