Question

(2x-3)(6x-1)(3x-2)(x-2)-7=0 solve the equation by partly factored polynomials​

Answer 1

Answer:

The required roots are

$\bold{\frac{1}{2},\frac{5}{3},\frac{13+i\sqrt{143}}{12},\frac{13-i\sqrt{143}}{12}}$

Step-by-step explanation:

(2x-3)(6x-1)(3x-2)(x-12)-7=0 solve the equation by partly factored polynomials

$(2x-3)(6x-1)(3x-2)(x-12)-7=0$

Rearranging we get

$(2x-3)(3x-2)(6x-1)(x-12)-7=0$

$(6x^2-13x+6)(6x^2-13x+12)-7=0$

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

$6x^2-13x=-13$

$6x^2-13x+13=0$

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

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

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

Answer 2

Answer:

Step-by-step explanation:

There is a mistake in the solution. (6x-1)(x-12) is not equal to 6x^2-13x+12.