Question : The LCM of $x^2-8x + 15$ and $x^2-5x + 6$ is:
Option 1: $(x-2)(x-3)^2(x-5)$
Option 2: $(x + 5) (x + 2) (x + 3)$
Option 3: $(x + 5) (x - 2) (x - 3)$
Option 4: $(x - 5) (x - 2) (x - 3)$
Correct Answer: $(x - 5) (x - 2) (x - 3)$
Solution : Given: The equations are $x^2- 8x + 15$ and $x^2- 5x + 6$. LCM is the product of every prime factor that is present in the numbers to the greatest power. The factors of $x^2-8x+15$ is given as, $=x^2-3x-5x+15$ $=(x-3)-5(x-3)$ $=(x-3)(x-5)$ The factors of $x^2-5x+6$ is given as, $=x^2-3x-2x+6$ $=(x-3)-2(x-3)$ $=(x-3)(x-2)$ LCM of $(x^2- 8x + 15$, $x^2- 5x + 6)=(x-5)(x-2)(x-3)$ Hence, the correct answer is $(x-5)(x-2)(x-3)$.
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Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : If $x^2-2\sqrt{10}x+1=0$, what is the value of $(x-\frac{1}{x})$?
Option 1: $4$
Option 2: $6$
Option 3: $3$
Option 4: $5$
Question : If $x^2-8x+1=0$, what is the value of $(x^2+\frac{1}{x^2})$?
Option 1: $18$
Option 2: $34$
Option 3: $40$
Option 4: $62$
Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Option 1: 4
Option 2: 5
Option 3: 6
Option 4: 8
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