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3 years ago

Solve.

Mathematics

1 answer:

Ray Of Light [21]3 years ago

5 0

Step-by-step explanation:

let d+e=2 be equation 1

let d-e=4 be equation 2

lets make d the subject in equa 1

d = 2-e equation 3

put equation 3 in 2

2-e=4

e=2-4

e=-2

put e=-2 in equa 1

d-2=2

d=4

e is -2 and d is 4

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First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

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Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

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$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

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$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

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$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

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2 years ago

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\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261

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