Question

Solve by using quadratic formula: (a)x - (1/x) = 5 (b) 1/x 1/(x 4) = 1/2 (c) 1/x 1/(x - 6) = 1/3​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given:

$a)x - \frac{1}{x} = 5$

$b) \frac{1}{x} - \frac{1}{x + 4} = \frac{1}{2}$

$c) \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{3}$

To find: Solve by using quadratic formula.

Solution:

Tip:

Quadratic formula

$\boxed{\bold{\pink{x_{1,2} = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}}}}$

$a)x - \frac{1}{x} = 5$

Step 1: Take LCM in LHS and cross multiply

$\frac{ {x}^{2} - 1}{x} = 5$

or

${x}^{2} - 1 = 5x$

or

${x}^{2} - 5x - 1 = 0$

Step 2: Apply Quadratic formula

$x_{1,2} = \frac{ 5± \sqrt{ {( - 5)}^{2} - 4(1)( - 1)} }{2}$

$x_{1,2} = \frac{5 ± \sqrt{29} }{2}$

or

$\bold{x_1= \frac{5 + \sqrt{29} }{2} } \\ \\\bold{ x_2 = \frac{5 - \sqrt{29} }{2} }$

$b) \frac{1}{x} - \frac{1}{x + 4} = \frac{1}{2}$

Step 1:Take LCM and solve to convert into standard quadratic equation

$\frac{x + 4 - x}{x(x + 4)} = \frac{1}{2}$

or

$\frac{ 4}{ {x}^{2} + 4x } = \frac{1}{2}$

or

${x}^{2} + 4x - 8 = 0$

Step 2: Apply Quadratic formula

$x_{1,2} = \frac{ - 4 ± \sqrt{16 + 32} }{2}$

$x_{1,2} = \frac{ - 4 ± \sqrt{48} }{2}$

or

$x_{1,2} = \frac{ - 4 ± 4 \sqrt{3} }{2}$

or

$x_{1,2} = - 2 ±2 \sqrt{3}$

or

$\bold{x_{1} = - 2 +2 \sqrt{3}}$

$\bold{x_{2} = - 2 -2 \sqrt{3}}$

$c) \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{3}$

Step 1:Take LCM and solve to convert into standard quadratic equation

$\frac{x - 6 + x}{x(x - 6)} = \frac{1}{3}$

or

$3(2x - 6) = {x}^{2} - 6x$

or

$6x - 18 - {x}^{2} + 6x = 0$

or

${x}^{2} - 12x + 18 = 0$

Step 2: Apply Quadratic formula

$x_{1,2} = \frac{12 ± \sqrt{144 - 72} }{2}$

or

$x_{1,2} = \frac{12 ± 6 \sqrt{2} }{2}$

or

$x_{1,2} = {6 ±3 \sqrt{2} }$

or

$\bold{x_1 = {6 + 3 \sqrt{2} }}$

$\bold{x_2= {6 - 3 \sqrt{2} }}$

Final answer:

$a)\pink{x = \frac{5 + \sqrt{29} }{2}} \\ \\ \pink{x = \frac{5 - \sqrt{29} }{2} }$

$b)\green{x = - 2 + 2 \sqrt{3} } \\ \\ \green{x = - 2 - 2 \sqrt{3}}$

$c)\red{x = {6 + 3 \sqrt{2} }}$

$\red{x = {6 - 3 \sqrt{2}} }$

Hope it helps you.

Note*: Questions are corrected.

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For what value of k does the Quadratic Equation X²-4kx+4k = 0 have equal positive integer solutions

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