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2 years ago

16x² +8-7x-3x2 - 13x

Mathematics

1 answer:

liq [111]2 years ago

5 0

Answer:

$13x^{2}$ -20x+8

Step-by-step explanation:

Step 1: rearrange the equation

16x^2 - 3x^2 - 7x - 13x +8

Step 2: solve the equation

16x^2 - 3x^2 = 13x^2

-7x - 13x = -20x

(the +8 will stay the same)

so when we put it all together it gives us

13x^2 - 20x + 8

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Step-by-step explanation:

(456/12) / (12/2)

lets solve the brackets

38/6

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$\large\underline{\sf{Solution-}}$

<u>Let us assume that:</u>

$\sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }$

We can also write it as:

$\sf \longmapsto x = \sqrt{6 + x }$

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$\sf \longmapsto {x}^{2} =6 + x$

$\sf \longmapsto {x}^{2} - x - 6 =0$

By splitting the middle term:

$\sf \longmapsto {x}^{2} - 3x + 2x - 6 =0$

$\sf \longmapsto x(x - 3) + 2(x - 3 ) =0$

$\sf \longmapsto (x+ 2)(x - 3 ) =0$

<u>Therefore:</u>

$\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}$

$\sf \longmapsto x = - 2,3$

<u>But x cannot be negative. </u>

$\sf \longmapsto x = 3$

Therefore, the value of the expression is 3.

$\large\underline{\sf{Verification-}}$

Given:

$\sf\longmapsto x=3$

We can also write it as:

$\sf\longmapsto x = \sqrt{9}$

$\sf\longmapsto x = \sqrt{6+3}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{9}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}$

This pattern will continue.

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}$

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