KnowledgeBoat Logo
|

Mathematics

Solve :

8x+3−2=32−x\dfrac{8}{x + 3} - 2 = \dfrac{3}{2 - x}

Quadratic Equations

7 Likes

Answer

Solving :

⇒8x+3−2=32−x⇒8−2(x+3)x+3=32−x⇒8−2x−6x+3=32−x⇒2−2xx+3=32−x⇒(2−2x)(2−x)=3(x+3)⇒4−2x−4x+2x2=3x+9⇒2x2−6x+4=3x+9⇒2x2−6x−3x+4−9=0⇒2x2−9x−5=0⇒2x2−10x+x−5=0⇒2x(x−5)+1(x−5)=0⇒(2x+1)(x−5)=0⇒2x+1=0 or x−5=0⇒x=−12 or x=5.\Rightarrow \dfrac{8}{x + 3} - 2 = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{8 - 2(x + 3)}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{8 - 2x - 6}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow \dfrac{2 - 2x}{x + 3} = \dfrac{3}{2 - x} \\[1em] \Rightarrow (2 - 2x)(2 - x) = 3(x + 3) \\[1em] \Rightarrow 4 - 2x - 4x + 2x^2 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x + 4 = 3x + 9 \\[1em] \Rightarrow 2x^2 - 6x - 3x + 4 - 9 = 0 \\[1em] \Rightarrow 2x^2 - 9x - 5 = 0 \\[1em] \Rightarrow 2x^2 - 10x + x - 5 = 0 \\[1em] \Rightarrow 2x(x - 5) + 1(x - 5) = 0 \\[1em] \Rightarrow (2x + 1)(x - 5) = 0 \\[1em] \Rightarrow 2x + 1 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow x = -\dfrac{1}{2} \text{ or } x = 5.

Hence, x = −12-\dfrac{1}{2} or 5.

Answered By

2 Likes

Related Questions