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3 years ago

Please help me simplify this equation!

Mathematics

1 answer:

vladimir1956 [14]3 years ago

3 0

X^2(x+4) x 3x^2(x-1)
(X^3+4x^2) x (3x^3 -3x^2)
3x^6 -3x^5+12x^5 -12x^4
3x^6 +9x^5-12x^4
I think that’s how you do it

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Let A = {1, 2, 3, 4, 5} and B = {2, 4}. What is A ∩ B?

Harlamova29_29 [7]

$A\cap B=\{x:x\in A \wedge x\in B\}$

$A\cap B=\{2,4\}$

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<img src="https://tex.z-dn.net/?f=%20%20%5Csf%20%5Chuge%7B%20question%20%5Chookleftarrow%7D" id="TexFormula1" title=" \sf \huge

BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

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$\bull\sf Apply\: Squares$

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$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

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$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

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$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

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Sever21 [200]

Answer:

$\{x=\frac{19-\sqrt{409} }{2}\ , \ x=\frac{19+\sqrt{409} }{2}\}$

Step-by-step explanation:

x² = 19x + 12

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b² - 4ac = (-19)² - 4×1×(-12) = 409

The discriminant is positive ,then the equation has two solutions.

$x=\frac{19-\sqrt{409} }{2} \ \ or\ \ x=\frac{19+\sqrt{409} }{2}$

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