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STALIN [3.7K]
3 years ago
5

Which of the following is an extraneous solution of (45-3x)^1/2=x-9?

Mathematics
1 answer:
Iteru [2.4K]3 years ago
3 0

Answer:

C. x =3

Step-by-step explanation:

Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.

Let's solve the equation first

\sqrt{45-3x} = x-9\\Taking\ square\ on\ both\ sides\\{(\sqrt{45-3x})}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\\sqrt{45-3(3)} = 3-9\\\sqrt{45-9} = -6\\\sqrt{36}= -6\\6\neq -6\\For x=12\\\sqrt{45-3(12)} = 12-9\\\sqrt{45-36} = 6\\\sqrt{36}= 6\\6=6

Hence, we can conclude that x=3 is an extraneous solution of the equation ..

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3 years ago
Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.
Mice21 [21]

Answer:

<em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

Step-by-step explanation:

Making factors of 8y^{6}+ 144y^{5}+ 640y^{4}

Taking 8y^{4} common:

\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)

Using <em>factorization</em> method:

\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times  4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)

Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)

Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

We know the relation between <em>LCM, HCF</em> of the two numbers <em>'p' , 'q'</em> and the <em>numbers</em> themselves as:

HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

Hence, <em>LCM</em> = 8y^{4}(y+ 10)^{2}(y + 8)

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