Question

Solve the following quadratic equation using the formula method:
7x/3x-4+2x/6-x=10

Answer 1

We reduce fractions to a common denominator and get

$$\frac {7x} {3x-4} +\frac {2x} {6-x} - 10=0$$

$$\frac {7x(6-x)} {(3x-4)(6-x)} +\frac {2x(3x-4)} {(6-x)(3x-4)} - \frac {10(6-x)(3x-4)} {(6-x)(3x-4)}=0$$

$$\frac {42x-7x^2+6x^2-8x+30x^2+180x-240+40x}{(6-x)(3x-4)}=0$$

$$\frac {29x^2 - 186x + 240}{(6-x)(3x-4)}=0$$

A fraction is zero if its numerator is zero.

$$29x^2 - 186x + 240=0$$

Solve the quadratic equation. Find the discriminant

$$D=186^2-4\times 29 \times 240=34596-27840=6756$$

Therefore the roots of the equation are equal

$$x_1=\frac {186-\sqrt {6756}} {2\times 29}= \frac {93-\sqrt {1689}} { 29}$$

$$x_2=\frac {186+\sqrt {6756}} {2\times 29}= \frac {93+\sqrt {1689}} { 29}$$

Answer.

$$x_1= \frac {93-\sqrt {1689}} { 29}$$

$$x_2= \frac {93+\sqrt {1689}} { 29}$$