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Lesechka [4]
3 years ago
10

Solve the following quadratic equation by factorisation method 3 x square - 2 root 6 × + 2 =0​

Mathematics
1 answer:
frozen [14]3 years ago
6 0
3x^2 -2root(6)x + 2 = 0

(root(3)x)^2 -2[root(3)x]root(2)+ (root(2))^2 = 0

let a = root(3)x and b = root(2)

a^2 -2ab + b^2 = 0

(a - b)^2 = 0

by the null factor rule,

a - b = 0

sub back in a and b

root(3)x = root(2)

x = root(3)/root(3)
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aksik [14]

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5 0
4 years ago
Read 2 more answers
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Ede4ka [16]
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6 0
4 years ago
Write each sum or difference as a product with positive arguments -4(sin4x - sin2x) A) 8cos 3xsin x B) 8sin xsin 3x C) -8cos 3xs
pshichka [43]

Answer:

C) -8cos 3xsin x

Step-by-step explanation:

To express -4(sin4x - sin2x)  as a product, we use the formula sinA - sinB = 2cos[(A + B)/2]sin[(A - B)/2.

Comparing sin4x - sin2x with sinA - sinB, A = 4x and B = 2x.

Substituting these into the equation, we have

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So, -4(sin4x - sin2x) = -8cos3xsinx

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5 0
3 years ago
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Aloiza [94]

Answer:

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5 0
3 years ago
8. Find all the real fourth roots of 256
mezya [45]
X^4=256
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or x^2-16=0
or x^2-4^2=0
(x+4)(x-4)=0
either x+4=0,x=-4
or x-4=0
x=4
7 0
3 years ago
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