{"id":2928,"date":"2025-07-26T02:05:00","date_gmt":"2025-07-26T06:05:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=2928"},"modified":"2025-08-18T12:54:20","modified_gmt":"2025-08-18T16:54:20","slug":"summer-school-exam-1-2025","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=2928","title":{"rendered":"HCS Summer School Exam 1 – 2025"},"content":{"rendered":"\n

Problem1:<\/strong> Let $x_{1}$ and $x_{2}$ be the root of $x^2-7x-9=0$. Find the value of $|x_{1}-x_{2}|$. Solution<\/a><\/p>\n\n\n\n

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$$|x_{1}-x_{2}|=\\dfrac{\\sqrt{b^2-4ac}}{|a|}=\\dfrac{\\sqrt{(-7)^2-4\\cdot(-9)}}{|1|}=\\sqrt{85}$$<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 2: <\/strong>Find all solutions of $\\sqrt{x+10}-\\dfrac{6}{\\sqrt{x+10}}=5$. Solution<\/a><\/p>\n\n\n\n

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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 $x=\\boxed{26}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 3:<\/strong> Find all possible values of $(a, b)$, so that $2a+b=12$, and $ab=3$. Solution<\/a><\/p>\n\n\n\n

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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 $\\boxed{(3-\\dfrac{\\sqrt{30}}{2}, 6+\\sqrt{30})}$<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 4:<\/strong> $4x^2-24x+c$ is a perfect square for all integer $x$. Find the value of $c$. Solution<\/a><\/p>\n\n\n\n

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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 $c=b^2=\\boxed{36}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 5:<\/strong> Find all $z$ such that $9^{z-1}-3^{z-1}-2=0$. Solution<\/a><\/p>\n\n\n\n

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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 $z=\\boxed{\\log_{3}6}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 6:<\/strong> Find all solutions to the equation $x+\\sqrt{x-2}=4$. Solution<\/a><\/p>\n\n\n\n

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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 $x=\\boxed{3}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 7:<\/strong> If $\\dfrac{a+b}{a}=\\dfrac{b}{a+b}$, then (A) <\/strong>No solutions for $a$ and $b$. (B)<\/strong> $a$ and $b$ cannot be both real. (C)<\/strong> Both $a$ and $b$ are imaginary. (D)<\/strong> One of $a$ or $b$ is real, the other imaginary. (E) <\/strong>Both $a$ and $b$ are real. Solution<\/a><\/p>\n\n\n\n

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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 $$a=\\dfrac{-b\\pm\\sqrt{-3b^2}}{2},\\ b=\\dfrac{-a\\pm\\sqrt{-3a^2}}{2}$$<\/p>\n\n\n\n

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

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)<\/strong>.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 8: <\/strong>Find all solutions to the equation $\\dfrac{3^{x^2}}{27^x}=\\dfrac{1}{9}$ Solution<\/a><\/p>\n\n\n\n

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Transforming the equation as $$3^{x^2-3x}=3^{-2}$$, we have $$x^2-3x=-2$$. Therefore $x=\\boxed{1}$, or $x=\\boxed{2}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 9: <\/strong>How many solutions are there for equation $|x^2-5|=4$? Solution<\/a><\/p>\n\n\n\n

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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.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 10:<\/strong> Find the sum of all solutions for the equation $1+\\dfrac{2}{x}=x$ Solution<\/a><\/p>\n\n\n\n

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

Problem 11: <\/strong>Simplify $\\sqrt{53-8\\sqrt{15}}$. Solution<\/a><\/p>\n\n\n\n

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

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 $$x^2+15y^2=53,\\ 2xy=-8$$<\/p>\n\n\n\n

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

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 $\\sqrt{53-8\\sqrt{15}}=\\boxed{4\\sqrt{3}-\\sqrt{5}}$.<\/p>\n\n\n\n

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 $$3x^2+5y^2=53,\\ 2xy=-8$$<\/p>\n\n\n\n

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

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 $\\sqrt{53-8\\sqrt{15}}=\\boxed{4\\sqrt{3}-\\sqrt{5}}$.<\/p>\n\n\n\n<\/div>\n\n\n\n

Problem 12: <\/strong>Find all solutions to the equation $\\sqrt{x+\\sqrt{x+11}}+\\sqrt{x-\\sqrt{x+11}}=4$ Solution<\/a><\/p>\n\n\n\n

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

Simplifying it, we have $$2x+2\\sqrt{x^2-x-11}=16$$<\/p>\n\n\n\n

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

Squaring both side again, we have $$x^2-x-11=64-16x+x^2$$ Therefore $15x=75$, $x=5$. Verifying the result, we have $x=\\boxed{5}$. <\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Problem1: Let $x_{1}$ and $x_{2}$ be the root of $x^2-7x-9=0$. Find the value of $|x_{1}-x_{2}|$. Solution $$|x_{1}-x_{2}|=\\dfrac{\\sqrt{b^2-4ac}}{|a|}=\\dfrac{\\sqrt{(-7)^2-4\\cdot(-9)}}{|1|}=\\sqrt{85}$$ Problem 2: Find all solutions of $\\sqrt{x+10}-\\dfrac{6}{\\sqrt{x+10}}=5$. Solution Let $y=\\sqrt{x+10}$, we have $$y-\\dfrac{6}{y}=5$$ Multiplying $y$ on both side of the equation, we have … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2928"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=2928"}],"version-history":[{"count":95,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2928\/revisions"}],"predecessor-version":[{"id":5129,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/2928\/revisions\/5129"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=2928"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=2928"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=2928"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}