HCS Summer School Exam 1 – 2025

$$|x_{1}-x_{2}|=\dfrac{\sqrt{b^2-4ac}}{|a|}=\dfrac{\sqrt{(-7)^2-4\cdot(-9)}}{|1|}=\sqrt{85}$$

Let $y=\sqrt{x+10}$, we have $$y-\dfrac{6}{y}=5$$ Multiplying $y$ on both side of the equation, we have $$y^2-6=5y$$ Moving $5y$ to the left side of the above equation and factorizing it, we have $$(y-6)(y+1)=0$$ Therefore $y=6$, $y=-1$. Obviously, $y=-1$ is invalid as $\sqrt{x+10}>0$. Therefore, $\sqrt{x+10}=6$, which implies $x+10=36$. Therefore $x=\boxed{26}$.

Since $ab=3$, we have $2ab=6$. According to Vieta’s theorem, $2a$ and $b$ are the roots of equation $$x^2-12x+6=0$$. Solving the above equation, we have $$2a=6\pm\sqrt{30},\ b=6\mp\sqrt{30}$$ Therefore, there are two solutions of $(a, b)$, as $\boxed{(3+\dfrac{\sqrt{30}}{2}, 6-\sqrt{30})}$ and $\boxed{(3-\dfrac{\sqrt{30}}{2}, 6+\sqrt{30})}$

Let $(ax+b)^2=4x^2-24x+c$, we have $a^2x^2+2abx+b^2=4x^2-24x+c$. Therefore $a^2=4$, $2ab=-24$, and $b^2=c$. This results in $a=\pm 2$, $b=\mp 6$, and $c=b^2=\boxed{36}$.

Let $3^{z-1}=x$, we have $x^2-x-2=0$. Solving the equation, $x=2$, $x=-1$. Obviously $x=-1$ is invalid, as $3^{z-1}>0$. Therefore $3^{z-1}=2$, which implies $3^z=6$. Therefore $z=\boxed{\log_{3}6}$.

Let $y=\sqrt{x-2}$, we have $y^2+y-2=0$. $y=1$, $y=-2$. Obviously, $y=-2$ is invalid, as $\sqrt{x-2}\ge 0$. Therefore $\sqrt{x-2}=1$, which leads $x=\boxed{3}$.

Obviously $a\ne 0$, $a+b\ne 0$. Multiplying both side with $a(a+b)$, we have $(a+b)^2=ab$. i.e. $a^2+ab+b^2=0$. We have $$a=\dfrac{-b\pm\sqrt{-3b^2}}{2},\ b=\dfrac{-a\pm\sqrt{-3a^2}}{2}$$

If $b$ is real, then $\sqrt{-3b^2}$ is imaginary, and $a$ is imaginary. Similarly, if $a$ is real, then $b$ is imaginary.

If $b$ is imaginary, $a$ could be either real (for example, when $b=1+i\sqrt{3}$, or imaginary (for example, when $b=i$). Similarly if $a$ is imaginary, $b$ could be either real, or imaginary. Therefore the correct answer is (B).

Transforming the equation as $$3^{x^2-3x}=3^{-2}$$, we have $$x^2-3x=-2$$. Therefore $x=\boxed{1}$, or $x=\boxed{2}$.

Transforming the equation as $x^2-5=\pm 4$, we have either $x^2-5=4$, or $x^2-5=-4$, which leads to $x^2=9$, or $x^2=1$. Therefore $x=\pm 3$, or $x=\pm 1$. Therefore there are $\boxed{4}$ solutions.

Transforming the equation as $x^2-x-2=0$. According to Vieta’s theorem, the sum of the roots is $-\dfrac{b}{a}=-\dfrac{-1}{1}=\boxed{1}$.

The simplified form should be either $x+y\sqrt{15}$, or $x\sqrt{3}+y\sqrt{5}$.

If the answer is in the form of $x+y\sqrt{15}$, then $(x+y\sqrt{15})^2=x^2+15y^2+2xy\sqrt{15}$, which leads to $$x^2+15y^2=53,\ 2xy=-8$$

Solving the above equation, we have $$(x, y)=(\pm 4\sqrt{3}, \mp\dfrac{\sqrt{3}}{3})$$ $$(x, y)=(\pm\sqrt{5},\mp\dfrac{4\sqrt{5}}{5})$$

Verifying the above 4 solutions, only $(x,y)=(-\sqrt{5},\dfrac{4\sqrt{5}}{5})$ and $(x,y)=(4\sqrt{3},-\dfrac{\sqrt{3}}{3})$ are the valid answers, both of them lead to $\sqrt{53-8\sqrt{15}}=\boxed{4\sqrt{3}-\sqrt{5}}$.

If the answer is in the form of $x\sqrt{3}+y\sqrt{5}$, then $(x\sqrt{3}+y\sqrt{5})^2=3x^2+5y^2+2xy\sqrt{15}$, which leads to $$3x^2+5y^2=53,\ 2xy=-8$$

Solving the above equation, we have $$(x, y)=(\pm 4,\mp 1)$$ $$(x,y)=(\pm\dfrac{\sqrt{15}}{3},\mp\dfrac{4\sqrt{15}}{5})$$

Verifying the above solutions, only $(x,y)=(4,-1)$ and $(x,y)=(-\dfrac{\sqrt{15}}{3},\dfrac{4\sqrt{15}}{5})$ are valid answers, both of them lead to $\sqrt{53-8\sqrt{15}}=\boxed{4\sqrt{3}-\sqrt{5}}$.

Squaring both side, we have $$\left(\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}\ \right)^2=16$$

Simplifying it, we have $$2x+2\sqrt{x^2-x-11}=16$$

Dividing both side by 2 and moving $x$ to the right side, we have $$\sqrt{x^2-x-11}=8-x$$

Squaring both side again, we have $$x^2-x-11=64-16x+x^2$$ Therefore $15x=75$, $x=5$. Verifying the result, we have $x=\boxed{5}$.