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Quadratic Equations — Maths STD 10 — Question

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

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