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Manipulation of a linear equation

x 2+8 x-38 x-...

Question

$\frac{x^{2} + 8 x - 3}{8 x - 3} = \frac{x^{2} + 4 x + 4}{4 x + 4}$

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Solution

$\frac{x^{2} + 8 x - 3}{8 x - 3} = \frac{x^{2} + 4 x + 4}{4 x + 4} \Rightarrow \frac{x^{2} + 8 x - 3 - 8 x + 3}{8 x - 3} = \frac{x^{2} + 4 x + 4 - 4 x - 4}{4 x + 4} (Dividendo) \Rightarrow \frac{x^{2}}{8 x - 3} = \frac{x^{2}}{4 x + 4} For x = 0, the equation is satisfied . Thus, x = 0 is one of the solutions . When x \neq 0, x^{2} \neq 0 . ∴ \frac{1}{8 x - 3} = \frac{1}{4 x + 4} (Dividing by x^{2}) \Rightarrow 4 x + 4 = 8 x - 3 \Rightarrow - 4 x = - 7 \Rightarrow x = \frac{7}{4} Hence, either x = 0 or x = \frac{7}{4} is the solution .$

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Similar questions

Q. After solving this equation

\frac{x^{2} + 8 x - 3}{8 x - 3} = \frac{x^{2} + 4 x + 4}{4 x + 4}

x

\frac{7}{4}

m

. Then, the value of

m

$lim x \to \infty (\sqrt{(x^{2} + 8 x + 3)} - \sqrt{(x^{2} + 4 x + 3)}) =$

lim x \to \infty (\sqrt{x^{2} + 8 x + 3} - \sqrt{x^{2} + 4 x + 3})

Q. If the zeroes of p(x) are double the zeroes of the polynomial q(x) =

x^{2} - 4 x + 3

Q. Tick (✓) the correct answer:
(x + 4)(x + 4) = ?
(a) (x² + 16)
(b) (x² + 4x + 16)
(c) (x² + 8x + 16)
(d) (x² + 16x)

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Manipulation of a linear equation

Standard XII Mathematics

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