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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{1}{(\text{x}-1)(\text{x}-2)}+\frac{1}{(\text{x}-2)(\text{x}-3)}+\frac{1}{(\text{x}-3)(\text{x}-4)}=\frac{1}{6}$
✓

Answer

We have been given,
$\frac{1}{(\text{x}-1)(\text{x}-2)}+\frac{1}{(\text{x}-2)(\text{x}-3)}+\frac{1}{(\text{x}-3)(\text{x}-4)}=\frac{1}{6}$
$\frac{(\text{x}-3)(\text{x}-4)+(\text{x}-1)(\text{x}-4)+(\text{x}-1)(\text{x}-2)}{(\text{x}-1)(\text{x}-4)(\text{x}-2)(\text{x}-3)}=\frac{1}{6}$
$\frac{3(\text{x}^2-5\text{x}+6)}{(\text{x}^2-5\text{x}+4)(\text{x}^2-5\text{x}+6)}=\frac{1}{6}$
$18=x^2-5 x+4 $
$x^2-5 x-14=0 $
$x^2-7 x+2 x-14=0 $
$x(x-7)+2(x-7)=0 $
$(x+2)(x-7)=0$
Therefore,
$x + 2 = 0$
$x = -2$
$x = 7$
Hence, $x = -2$ or $x = 7$

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