The roots of x-4 x-4-x-4 x-4-103 area - 4b - 6c - 8d 2 - 3

(c) plusmn;8 Given: #160;x #160;+ #160;4x #160; #8722; #160;4 #160;+ #160;x #160; #8722; #160;4x #160;+ #160;4 #160;= #160;103 ...

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Standard X

Mathematics

Quadratic Equations

The roots of ...

Question

The roots of $\frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = \frac{10}{3}$ are
(a) ±4
(b) ±6
(c) ±8
(d) $2 \pm \sqrt{3}$

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Solution

(c) ±8
$Given: \frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = \frac{10}{3} \Rightarrow \frac{{(x + 4)}^{2} + {(x - 4)}^{2}}{(x - 4) (x + 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} + 8 x + 16 + x^{2} - 8 x + 16}{x^{2} - 16} = \frac{10}{3} \Rightarrow 3 (2 x^{2} + 32) = 10 (x^{2} - 16) \Rightarrow 6 x^{2} + 96 = 10 x^{2} - 160 ⇒ 4 x^{2} = 256 ⇒ x^{2} = 64 \Rightarrow x = \pm 8$

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If $\frac{4 x + 8}{5 x + 8} = \frac{5}{6}$ , then x = ?

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Question 10
Which of the following equations has no real roots?
(A) $x^{2} - 4 x + 3 \sqrt{2} = 0$
(B) $x^{2} + 4 x - 3 \sqrt{2} = 0$
(C) $x^{2} - 4 x - 3 \sqrt{2} = 0$
(D) $3 x^{2} - 4 \sqrt{3} x + 4 = 0$

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The roots of x+4x-4+x-4x+4=103 are (a) ±4 (b) ±6 (c) ±8 (d) 2±3

(c) ±8 Given: x + 4x − 4 + x − 4x + 4 = 103⇒ (x + 4)2 + (x − 4)2(x − 4)(x + 4) = 103⇒ x2 + 8x + 16 + x2 − 8x + 16x2 − 16 = 103⇒ 3(2x2 + 32) = 10(x2 − 16)⇒ 6x2 + 96 = 10x2 − 160⇒ 4x2 = 256⇒ x2 = 64⇒ x= ±8

The roots of $\frac{x + 4}{x - 4} + \frac{x - 4}{x + 4} = \frac{10}{3}$ are
(a) ±4
(b) ±6
(c) ±8
(d) $2 \pm \sqrt{3}$