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2 years ago

Need Help BadlyWhat is the solution

Mathematics

1 answer:

Kamila [148]2 years ago

5 0

Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8

The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.

After doing this, we realize that only -1 works, so the answer is x=-1

Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)

Best wishes:)

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Answer:

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Step-by-step explanation:

Given expression is $(7x^2+a^2)(x^2-3a^2)$

To find the multiplication of the given expression :

$(7x^2+a^2)(x^2-3a^2)$

By using the distributive property to solve the given expression as below

$(7x^2+a^2)(x^2-3a^2)=(7x^2)(x^2)+(7x^2)(-3a^2)+(a^2)(x^2)+(a^2)(-3a^2)$

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$=7x^{2+2}-21x^2a^2+a^2x^2-3a^{2+2}$ ( by using the property $a^m.a^n=a^{m+n}$ )

$=7x^4-21x^2a^2+a^2x^2-3a^4$

Adding the like terms in the above expression we get

$=7x^4-20x^2a^2-3a^4$

Therefore $(7x^2+a^2)(x^2-3a^2)=7x^4-20x^2a^2-3a^4$

Therefore the given multiplied expression is

$(7x^2+a^2)(x^2-3a^2)=7x^4-20x^2a^2-3a^4$

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