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4 years ago

Find the solution(s) to x2 – 14x + 49 = 0

Mathematics

2 answers:

MatroZZZ [7]4 years ago

7 0

Answer:

x=7

Step-by-step explanation:

$x^2 - 14x + 49 = 0$

Factor the left hand side of the equation

product is 49 and sum is -14

-7 times -7 is 49

-7 plus -7 is -14

so two factors are -7 and -7

$x^2 - 14x + 49 = 0$

$(x-7)(x-7)= 0$

Now we apply zero factor property

SEt each factor =0 and solve for x

x-7 =0 , x=7

Nitella [24]4 years ago

6 0

Add 49 to 0

2x-14x=49

then combine like terms

12x=49

x=49/12

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Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%7B%20%5Cinfty%20%7D%20%20%5Carcsin%20%5Clar

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Recall that over an appropriate domain,

$\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)$

Let $x=\frac1{n+1}$ and $y=\frac1{n+2}$ (these belong to the "appropriate domain", so the identity holds). We have

$\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}$

and

$\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}$

Then the sum telescopes, as

$\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}$

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