Elementary Algebra
 The degree f polynomial p(x) = x^{3} + 1 + 2x = 6x + 1/x is ?

 2
 4
 3
 5

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x^{3} + 1 + 2x =6x + 1/x
x^{3} + 1 + 2x = (6x^{2} + 1)/x
(x^{3} + 1 + 2x)x = 6x^{2} + 1
x^{4}  4x^{2}  6x^{2} 1 =0
x^{4}  4x^{2} + x  1 =0Correct Option: B
x^{3} + 1 + 2x =6x + 1/x
x^{3} + 1 + 2x = (6x^{2} + 1)/x
(x^{3} + 1 + 2x)x = 6x^{2} + 1
x^{4}  4x^{2}  6x^{2} 1 =0
x^{4}  4x^{2} + x  1 =0
Degree of polynomial is highest exponent degree term i.e.,4.
 If x = 1 + √2, then what is the value of x^{4}  4x^{3} + 4x^{2} ?

 1
 0
 1
 2

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x = 1 + √2
∴ x^{4}  4x^{3} + 4x^{2} = x^{2}(x^{2}  4x + 4)
= x^{2}(x  2)^{2}
= (1 + √2)^{2}(1 + √2  2)^{2}Correct Option: C
x = 1 + √2
∴ x^{4}  4x^{3} + 4x^{2} = x^{2}(x^{2}  4x + 4)
= x^{2}(x  2)^{2}
= (1 + √2)^{2}(1 + √2  2)^{2}
=(√2 + 1)^{2} (√2  1)^{2}
=[(√2)^{2}  (1)^{2}]^{2}
=(2  1)^{2} =1
 What is the solution of the equations x  y = 0.9 and 11(x + y)^{1} = 2 ?

 x = 3.2, y = 2.3
 x = 1, y = 0.1
 x = 2, y = 1.1
 x = 1.2, y = 0.3

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x  y = 0.9 ...(i)
and 11(x + y)^{1}=2
⇒ 11/ (x + y) = 2
⇒ 2(x + y) =11
Correct Option: A
x  y = 0.9 ...(i)
and 11(x + y)^{1}=2
⇒ 11/ (x + y) = 2
⇒ 2(x + y) =11
⇒ x + y = 11/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 3.2
and y = 2.3
 The product of two alternate odd integers exceeds three times the smaller by 12. What is the larger integer ?

 9
 7
 3
 5

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Let two alternate odd integers odd integers be (2x+1) and (2x+5).
Then according to the question,
(2x + 1) (2x + 5) = 3(2x + 1) + 12
⇒ (2x + 1)(2x + 5  3) = 12
⇒ 2x^{2} + 3x  5 = 0
Correct Option: B
Let two alternate odd integers odd integers be (2x+1) and (2x+5).
Then according to the question,
(2x + 1) (2x + 5) = 3(2x + 1) + 12
⇒ (2x + 1) (2x + 5  3) = 12
⇒ 2x^{2} + 3x  5 = 0
On solving this quadratic equation,we get
x = 1 and x = 5/2
x = 5/2 is not a integer ∴ x = 1
Then, larger integer = 2x + 5 = 2 x 1 + 5 = 7
 The sum of two numbers is 24 and their product is 143. The sum of their squares is

 296
 295
 290
 228

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Let the two numbers be P and Q.
According to given question,
Sum of two numbers = 24
∴ P + Q = 24
Product of two numbers = 143
and, PQ = 143
As we know the formula,
∴ P ^{2} + Q^{2} = (P + Q)^{2} – 2PQCorrect Option: C
Let the two numbers be P and Q.
According to given question,
Sum of two numbers = 24
∴ P + Q = 24 ...................... (1)
Product of two numbers = 143
and, PQ = 143 .......................... (2)
As we know the formula,
∴ P ^{2} + Q^{2} = (P + Q)^{2} – 2PQ
Put the value from the equation (1) and (2), We will get
∴ P ^{2} + Q^{2} = (24)^{2} – 2 × 143
∴ P ^{2} + Q^{2} = 576 – 286 = 290