Find the smallest value of \(a\) for which the equation \(\dfrac{x^2-x+a}{2x+3}=0\) has exactly one root.
Solution
We need \(x^2-x+a=0\), with \(x\ne-\dfrac32\). Exactly one admissible root can occur in two ways. If the quadratic has a double root, \(1-4a=0\), so \(a=\dfrac14\). Otherwise it has two roots but one is the excluded value \(-\dfrac32\). Substitution gives \(\dfrac94+\dfrac32+a=0\), hence \(a=-\dfrac{15}{4}=-3.75\). This is smaller than \(\dfrac14\).
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2
Task
One correct answer
Choose the equation whose root is \(2\).
Solution
Substitute \(x=2\). In \(\dfrac{3x-6}{x}=0\), the numerator is zero and the denominator is nonzero: \(\dfrac{3\cdot2-6}{2}=0\). Answer: D.
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3
Task
One correct answer
Solve the equation \(\dfrac{1}{2x}=\dfrac{1}{2-3x}\).
Solution
Domain restrictions: \(x\ne0\), \(x\ne\dfrac23\). Cross-multiplying gives \(2-3x=2x\), hence \(5x=2\) and \(x=0.4\). The value is admissible. Answer: D.
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4
Task
One correct answer
Solve the equation \(\dfrac{4}{|x|}=3\).
Solution
From \(\dfrac{4}{|x|}=3\), we get \(3|x|=4\), so \(|x|=\dfrac43\). Hence \(x=\pm\dfrac43\). Answer: E.
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5
Task
One correct answer
Choose the interval containing the root of \(\dfrac{3x-2}{x+1}=7\).
Solution
Domain restriction: \(x\ne-1\). We get \(3x-2=7(x+1)\), so \(3x-2=7x+7\), \(-9=4x\), and \(x=-\dfrac94=-2.25\). Thus the root lies in \((-\infty,-2)\). Answer: A.
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6
Task
Numeric answer
Solve \(\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}\). Enter the sum of the integer roots.
Solution
After clearing denominators, the numerator factors as \((x-3)(x-5)(x^2-18x+15)=0\). The roots are \(3\), \(5\), and \(9\pm\sqrt{66}\), and none is excluded by the denominators. The integer roots are \(3\) and \(5\), so their sum is \(8\).
Domain restrictions: \(x\ne-2,0,2\). Factor the denominators: \(x^2-4=(x-2)(x+2)\), \(x^2+2x=x(x+2)\), and \(x^2-2x=x(x-2)\). Multiplying by the common denominator gives \(2x+(x-4)(x-2)=x+2\). Thus \(x^2-5x+6=0\), so \(x=2\) or \(x=3\). Since \(2\) is excluded, \(x=3\). Answer: A.
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8
Task
One correct answer
Solve \(\dfrac{x^2+1}{x}+\dfrac{x}{x^2+1}=-2.5\).
Solution
Domain restriction: \(x\ne0\). Let \(t=\dfrac{x^2+1}{x}\). Then \(\dfrac{x}{x^2+1}=\dfrac1t\), so \(t+\dfrac1t=-\dfrac52\). Hence \(2t^2+5t+2=0\), giving \(t=-\dfrac12\) or \(t=-2\). The first value gives no real \(x\); from \(\dfrac{x^2+1}{x}=-2\) we get \((x+1)^2=0\), so \(x=-1\). Answer: B.
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9
Task
Numeric answer
Solve \(\dfrac{13x}{3x^2-5x+2}-\dfrac{12x}{3x^2+x+2}=5\). Enter the largest root.
Solution
After combining over a common denominator, the numerator factors as \((x-2)(3x-1)(15x^2+14x+10)\). The quadratic factor has no real roots. Thus the real candidates are \(x=2\) and \(x=\dfrac13\), both admissible. The largest root is \(2\).
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10
Task
Numeric answer
Solve \(\dfrac{12x}{3x^2-10x+1}-\dfrac1x=3x-6\). Enter the reciprocal of the product of all roots.
Solution
After clearing denominators, we obtain \((x-1)(3x-1)(3x^2-12x+1)=0\). All resulting roots are admissible. The product from the first two factors is \(1\cdot\dfrac13\), and by Vieta the product of the roots of \(3x^2-12x+1=0\) is \(\dfrac13\). Thus the total product is \(\dfrac19\), whose reciprocal is \(9\).
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11
Task
One correct answer
Which of the following intervals contains the root of \(\dfrac{x}{9-x}=\dfrac12\)?
Solution
Domain restriction: \(x\ne9\). From \(2x=9-x\), we get \(3x=9\), so \(x=3\). Therefore the root lies in \((2,5]\). Answer: D.
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12
Task
One correct answer
Choose the interval containing the root of \(\dfrac{x}{18-2x}=\dfrac14\).
Solution
Domain restriction: \(x\ne9\). Cross-multiplying gives \(4x=18-2x\), hence \(6x=18\) and \(x=3\). The root lies in \([0,4)\). Answer: C.
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13
Task
Numeric answer
Find all values of \(a\) for which \(\dfrac{x^2-ax+4}{x-5}=0\) has exactly one root. If there are several such values, enter their product.
Solution
Domain restriction: \(x\ne5\). The numerator gives \(x^2-ax+4=0\). Exactly one admissible root occurs when \(D=0\), so \(a=\pm4\), or when one of two roots is the excluded value \(5\): \(25-5a+4=0\), hence \(a=\dfrac{29}{5}=5.8\). Their product is \(-92.8\).
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14
Task
Numeric answer
Find all values of \(a\) for which \(\dfrac{x^2-4x+3}{3x-a}=\dfrac{42+3x-x^2}{3x-a}\) has exactly one root. If there are several, enter their sum.
Solution
Domain restriction: \(3x-a\ne0\). Equating numerators gives \(2x^2-7x-39=0\), so \(x_1=6.5\) and \(x_2=-3\). Exactly one root remains if the denominator vanishes at one of them: \(a=19.5\) or \(a=-9\). Their sum is \(10.5\).
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15
Task
Numeric answer
Two workers together can mow a plot in \(2\) hours \(6\) minutes. How many hours would the second worker need alone if this takes him \(4\) hours longer than the first worker?
Solution
Let the first worker need \(x\) hours, so the second needs \(x+4\). Since \(2\) h \(6\) min equals \(\dfrac{21}{10}\) h, \(\dfrac1x+\dfrac1{x+4}=\dfrac{10}{21}\). This gives \(5x^2-x-42=0\), whose positive solution is \(x=3\). Thus the second worker needs \(7\) hours.
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16
Task
Numeric answer
A bus travels from city \(A\) to city \(B\), \(150\) km apart. \(30\) minutes later a car leaves along the same route at \(1\dfrac15\) times the bus speed. How many hours does the car spend on the trip if they arrive at the same time?
Solution
Let the bus speed be \(x\) km/h, so the car speed is \(1.2x\). The travel-time difference is \(0.5\) h: \(\dfrac{150}{x}-\dfrac{150}{1.2x}=\dfrac12\). Thus \(x=50\) km/h, the car speed is \(60\) km/h, and its time is \(150/60=2.5\) h.
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17
Task
Numeric answer
The first pipe fills an empty pool \(40\) minutes faster than the second. How many minutes does the first pipe need alone if both pipes together fill the pool in \(21\) minutes?
Solution
Let the second pipe need \(x\) minutes, so the first needs \(x-40\). Then \(\dfrac1{x-40}+\dfrac1x=\dfrac1{21}\). This gives \(x^2-82x+840=0\), so \(x=12\) or \(70\). Only \(70\) fits the conditions, hence the first pipe needs \(30\) minutes.
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18
Task
Numeric answer
On a \(240\) km route, a train traveled \(10\) km/h slower than scheduled and arrived \(48\) minutes late. What was its scheduled speed in km/h?
Solution
Let the scheduled speed be \(x\) km/h, so the actual speed is \(x-10\). Since \(48\) minutes is \(\dfrac45\) h, \(\dfrac{240}{x-10}-\dfrac{240}{x}=\dfrac45\). This gives \(x^2-10x-3000=0\), with roots \(-50\) and \(60\). The valid speed is \(60\) km/h.
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19
Task
Numeric answer
A workshop planned to make \(240\) chairs in \(n\) days at a constant daily rate. The job was finished \(2\) days early by increasing the daily output by \(4\) chairs. Find \(n\).
Solution
Planned daily output: \(\dfrac{240}{n}\); actual output: \(\dfrac{240}{n-2}\). Thus \(\dfrac{240}{n-2}-\dfrac{240}{n}=4\). This simplifies to \(n^2-2n-120=0\), giving \(n=-10\) or \(12\). Hence \(n=12\).
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20
Task
Numeric answer
A boat travels \(18\) km upstream in half the time it takes to travel \(48\) km downstream. Find the boat’s speed in still water (km/h) if the current speed is \(2.5\) km/h.
Solution
Let the boat’s still-water speed be \(x\) km/h. Downstream time is \(\dfrac{48}{x+2.5}\), upstream time is \(\dfrac{18}{x-2.5}\). The condition gives \(\dfrac{48}{x+2.5}=2\cdot\dfrac{18}{x-2.5}\). Hence \(4(x-2.5)=3(x+2.5)\), so \(x=17.5\).
