Question
Solve equation using factorisation method:
$\frac{x-3}{x+3}+\frac{x+3}{x-3}=2 \frac{1}{2}$
$\frac{x-3}{x+3}+\frac{x+3}{x-3}=2 \frac{1}{2}$
✓
Answer
$\frac{x-3}{x+3}+\frac{x+3}{x-3}=2 \frac{1}{2}$
$\Rightarrow \frac{( x -3)^2+( x +3)^2}{( x +3)( x -3)}=\frac{5}{2}$
$\Rightarrow \frac{ x ^2-6 x +9+ x ^2+6 x +9}{ x ^2-9}=\frac{5}{2}$
$\Rightarrow 2\left(2 x^2+18\right)=5\left(x^2-9\right)$
$\Rightarrow 4 x^2+36=5 x^2-45$
$\Rightarrow x^2-81=0$
$\Rightarrow x^2-9^2=0$
$\Rightarrow (x + 9) (x - 9) = 0$
If $x +9 = 0$ or $x - 9 = 0$
then $x = -9$ or $x = 9$
$\Rightarrow \frac{( x -3)^2+( x +3)^2}{( x +3)( x -3)}=\frac{5}{2}$
$\Rightarrow \frac{ x ^2-6 x +9+ x ^2+6 x +9}{ x ^2-9}=\frac{5}{2}$
$\Rightarrow 2\left(2 x^2+18\right)=5\left(x^2-9\right)$
$\Rightarrow 4 x^2+36=5 x^2-45$
$\Rightarrow x^2-81=0$
$\Rightarrow x^2-9^2=0$
$\Rightarrow (x + 9) (x - 9) = 0$
If $x +9 = 0$ or $x - 9 = 0$
then $x = -9$ or $x = 9$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the given Figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is the radius of the circumcircle of A ABC, whose center is O.
Find the ratio in which the point $R ( 1, 5)$ divides the line segment joining the points $S (-2, -1)$ and $T (5, 13).$
→Solve the following equations for x and give, in each case, your answer correct to 3 decimal places
$2x^2 + 11x + 4= 0$
→$2x^2 + 11x + 4= 0$
Find the values of a and b) if [2a + 3b a - b] = [19 2].
Prove that : $\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta^{\prime}$
→The surface area of a sphere is $2464\ cm ^2$ ,find its volume
→In fig., PT is a tangent to the circle at T and PAB is a secant to the same circle. If $PB = 9cm$ and $AB = 5 cm$, find PT.
→Prove.
$\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2}=\frac{1-\sin A}{1+\sin A}$
→$\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2}=\frac{1-\sin A}{1+\sin A}$
In following fig. ABC is an equilateral triangle . A circle is drawn with centre A so that ot cuts AB and AC at M and N respectively. Prove that BN = CM.
Use the substitution $y= 2x +3$ to solve for x, if $4(2x+3)^2 – (2x+3) – 14 =0.$
→