Question : If $(x + 1)^2 + (x + 2)^2 = 16$, then what is the value of $4x^2 + 12x + 40$?
Option 1: 52
Option 2: 62
Option 3: 56
Option 4: 74
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Correct Answer: 62
Solution : $(x + 1)^2 + (x + 2)^2 = 16$ $⇒x^2+2x+1+x^2+4x+4=16$ $⇒2x^2+6x+5=16$ $⇒2x^2+6x=11$ Adding 20 on both sides, we get, $⇒2x^2+6x+20=11+20$ Multiplying 2 on both sides, we get, $⇒2(2x^2+6x+20)=2(31)$ $⇒4x^2 + 12x + 40=62$ Hence, the correct answer is 62.
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Question : If $4(x+5)-3>6-4x\geq x-5$, then the value of $x$ is:
Option 1: –2
Option 2: –1
Option 3: 3
Option 4: 4
Question : If $4x^2+y^2 = 40$ and $xy=6$, then the value of $2x-y$ is:
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 2
Question : If $x+\frac{1}{x}=2 K$, then what is the value of $x^4+\frac{1}{x^4}$?
Option 1: $16 {K}^4-16 {K}^2-1$
Option 2: $8 {K}^4+4 {K}^2-1$
Option 3: $16 {K}^4-16 {K}^2+2$
Option 4: $16 {K}^4-4 {K}^2-1$
Question : If $x+\frac{1}{x}=3$, then the value of $\frac{3x^{2}-4x+3}{x^{2}-x+1}$ is:
Option 1: $\frac{4}{3}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{5}{3}$
Question : If $x^2+\frac{1}{x^2}=4$, then what is the value of $x^4+\frac{1}{x^4}$?
Option 1: 16
Option 2: 14
Option 3: 12
Option 4: 20
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