Difficult questions on quadratic equations for 0
Answers
Answer:
the question:
Solve the equation \displaystyle \frac{5}{2-x}+\frac{x-5}{x+2}+\frac{3x+8}{x^2-4}=0
2−x
5
+
x+2
x−5
+
x
2
−4
3x+8
=0. In the answer box, write the roots separated by a comma
answer:
The equation is defined for x, such that \displaystyle x-2 \ne 0; x+2 \ne 0; x^2-4 \ne 0x−2
=0;x+2
=0;x
2
−4
=0, which yield us \displaystyle x \ne \pm 2x
=±2.
\displaystyle \frac{5}{2-x}+\frac{x-5}{x+2}+\frac{3x+8}{x^2-4}=0
2−x
5
+
x+2
x−5
+
x
2
−4
3x+8
=0
\displaystyle \frac{5(x+2)}{(2-x)(x+2)}+\frac{(x-5)(2-x)}{(x+2)(2-x)}+\frac{3x+8}{x^2-4}=0
(2−x)(x+2)
5(x+2)
+
(x+2)(2−x)
(x−5)(2−x)
+
x
2
−4
3x+8
=0
Have in mind that \displaystyle \frac{3x+8}{x^2 - 4}=-\frac{3x+8}{4-x^2}
x
2
−4
3x+8
=−
4−x
2
3x+8
\displaystyle \frac{5x+10}{4-x^2}+\frac{2x-10-x^2+5x}{4-x^2}-\frac{3x+8}{4-x^2}=0
4−x
2
5x+10
+
4−x
2
2x−10−x
2
+5x
−
4−x
2
3x+8
=0. We don't need the denominator anymore.
\displaystyle 5x+10+7x-10-x^2-3x-8=05x+10+7x−10−x
2
−3x−8=0
\displaystyle x^2-9x+8=0x
2
−9x+8=0
\displaystyle x^2-8x-x+8=0x
2
−8x−x+8=0
\displaystyle x(x-8)-(x-8)=0x(x−8)−(x−8)=0
\displaystyle (x-1)(x-8)=0(x−1)(x−8)=0, so the roots are x=1 and x=8. The equation is defined for them, so they are both solutions.