Choose the interval containing the solution of \(\sqrt{1-x}=4\).
Solution
Squaring gives \(1-x=16\), hence \(x=-15\). The number \(-15\) lies in \((-20;-10)\). Answer: A.
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2
Task
One correct answer
How many solutions does \(3\sqrt[8]{2x-9}-\sqrt{4-x}=1\) have?
Solution
We need both \(2x-9\ge0\) and \(4-x\ge0\). Thus \(x\ge\frac92\) and \(x\le4\), which is impossible. The equation has no solutions. Answer: A.
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3
Task
One correct answer
How many solutions does \(\sqrt[3]{x-3}=\sqrt{-2-x}\) have?
Solution
The square root requires \(x\le-2\). Then \(x-3\lt0\), so \(\sqrt[3]{x-3}\lt0\), while \(\sqrt{-2-x}\ge0\). Equality is impossible, so there are no solutions. Answer: C.
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4
Task
Numeric answer
Solve the equation \(\sqrt[3]{x+7}-\sqrt{x+3}=0\).
Solution
Domain: \(x\ge-3\). Let \(t=\sqrt{x+3}\ge0\), so \(x=t^2-3\). Then \(\sqrt[3]{t^2+4}=t\), hence \(t^3=t^2+4\). Thus \((t-2)(t^2+t+2)=0\). The only nonnegative root is \(t=2\), giving \(x=1\).
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5
Task
Numeric answer
Solve \(\sqrt[3]{x+3}+\sqrt{5-x}=2\). Enter the sum of all solutions.
Solution
Let \(a=\sqrt[3]{x+3}\) and \(b=\sqrt{5-x}\ge0\). Then \(a+b=2\) and \(a^3+b^2=8\). Substituting \(b=2-a\) gives \(a^3+a^2-4a-4=0\), so \((a+1)(a-2)(a+2)=0\). Hence \(x=-4,5,-11\), whose sum is \(-10\).
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6
Task
Numeric answer
Solve \(\sqrt{x-2}\sqrt{2x+1}=\sqrt3\). If there is one solution, enter it; if there are several, enter their product; if there are none, enter \(100\).
Solution
Domain: \(x\ge2\). Squaring gives \((x-2)(2x+1)=3\), so \(2x^2-3x-5=0\). The roots are \(x_1=2.5\) and \(x_2=-1\); the latter is outside the domain. Answer: \(2.5\).
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7
Task
One correct answer
Which interval contains the solution of \(\sqrt[3]{2x}=-3\)?
Solution
Cubing gives \(2x=-27\), so \(x=-13.5\). This lies in \((-20;-10)\). Answer: B.
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8
Task
One correct answer
If \(\sqrt{x}+y=5\) and \(2\sqrt{x}-y=7\), then \(y\) equals:
Solution
Adding the equations gives \(3\sqrt{x}=12\), so \(\sqrt{x}=4\). From the first equation, \(y=5-4=1\). Answer: E.
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9
Task
Numeric answer
The graph of \(y=\sqrt{2x^2+x+1}\) passes through \((x_0,4)\), where \(x_0\gt0\). Find \(x_0\).
Solution
Substituting the point gives \(4=\sqrt{2x_0^2+x_0+1}\). Squaring yields \(2x_0^2+x_0-15=0\), with roots \(2.5\) and \(-3\). Since \(x_0\gt0\), we get \(x_0=2.5\).
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10
Task
One correct answer
Choose the interval containing the solution of \(\sqrt{6-4x}=4\).
Solution
Squaring gives \(6-4x=16\). Hence \(-4x=10\), so \(x=-2.5\). This lies in \([-3;-1)\). Answer: A.
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11
Task
Numeric answer
Solve \(x-5+\sqrt{2x^2-14x+13}=0\). If the equation has several roots, enter their product.
Solution
We have \(\sqrt{2x^2-14x+13}=5-x\), so \(x\le5\). Squaring gives \(2x^2-14x+13=(5-x)^2\), hence \(x^2-4x-12=0\). The candidates are \(6\) and \(-2\); only \(-2\) satisfies \(x\le5\).
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12
Task
Numeric answer
Solve \((x^2-9)\sqrt{-15+8x-x^2}=0\). Enter the sum of the roots.
Solution
Domain: \(-15+8x-x^2\ge0\), so \(x\in[3;5]\). The product is zero when \(x^2-9=0\) or \(-15+8x-x^2=0\). Within the domain the roots are \(3\) and \(5\), whose sum is \(8\).
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13
Task
One correct answer
Solve \(4\sqrt{x}=1\).
Solution
\(\sqrt{x}=\dfrac14\). Squaring gives \(x=\dfrac1{16}\). Answer: E.
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14
Task
One correct answer
Solve \(\sqrt{3x+7}=x-7\).
Solution
The right-hand side must be nonnegative, so \(x\ge7\). Squaring gives \(3x+7=(x-7)^2\), hence \(x^2-17x+42=0\). The candidates are \(3\) and \(14\); only \(14\) satisfies \(x\ge7\). Answer: B.
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15
Task
One correct answer
Solve \(\sqrt{x+3}+\sqrt{4x-23}-\sqrt{x+10}=0\).
Solution
Domain: \(x\ge\dfrac{23}{4}\). Rearranging and squaring gives \(\sqrt{(x+3)(4x-23)}=15-2x\), so also \(x\le7.5\). Squaring again yields \(49x=294\), hence \(x=6\). Answer: A.
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16
Task
One correct answer
Solve \(2x^{\frac12}-5x^{\frac14}=3\).
Solution
Let \(t=x^{\frac14}\), where \(t\ge0\). Then \(x^{\frac12}=t^2\), so \(2t^2-5t-3=0\). Thus \(t=3\) or \(t=-\dfrac12\). Only \(t=3\) is admissible, hence \(x=81\). Answer: C.
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17
Task
Numeric answer
For what greatest negative value of the parameter \(a\) does \(\sqrt[4]{|x|-1}-2x=a\) have exactly one root?
Solution
For negative \(a\), roots can occur only on the branch \(x\ge1\), where \(f(x)=(x-1)^{1/4}-2x\). The greatest negative value giving one intersection is the maximum of this branch. From \(f'(x)=\dfrac{1}{4(x-1)^{3/4}}-2=0\), we get \(x=\dfrac{17}{16}\). Thus \(a=f\!\left(\dfrac{17}{16}\right)=-\dfrac{13}{8}=-1.625\).
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18
Task
Numeric answer
For what smallest value of \(a\) does \(\sqrt{x-2+2\sqrt{x-3}}+(14-2a)\sqrt[4]{x-3}+32=6a\) have at least one root?
Solution
Domain: \(x\ge3\). Let \(t=\sqrt[4]{x-3}\ge0\). Then \(\sqrt{x-3}=t^2\) and \(\sqrt{x-2+2\sqrt{x-3}}=t^2+1\). The equation becomes \(t^2+(14-2a)t+33-6a=0\), or \((t+3)(t+11-2a)=0\). Since \(t=-3\) is inadmissible, we need \(2a-11\ge0\). Thus the smallest value is \(a=5.5\).
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19
Task
Numeric answer
For what smallest integer value of the parameter \(a\) does \(\sqrt{2x+15}\left(\sqrt{x^2+18x+81}-\sqrt{x^2-10x+25}\right)=a\sqrt{2x+15}\) have exactly two distinct roots?
Solution
Domain: \(x\ge-7.5\). We have \(\sqrt{x^2+18x+81}=|x+9|\) and \(\sqrt{x^2-10x+25}=|x-5|\). Thus \(x=-7.5\) is a root for every \(a\), while any additional root satisfies \(|x+9|-|x-5|=a\). On the domain this function equals \(2x+4\) for \(-7.5\le x\le5\) and \(14\) for \(x\ge5\). Exactly one additional root occurs when \(-11\lt a\lt14\). The smallest integer is \(-10\).
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20
Task
Numeric answer
Find the smallest integer value of the parameter \(a\) for which \(\sqrt{x^2-5x}+\sqrt{x^2-9x+20}=\sqrt{a}\,\sqrt{x-5}\) has two roots.
Solution
The domain gives \(x\ge5\) and \(a\ge0\). The value \(x=5\) is a root for every admissible \(a\). For \(x\gt5\), divide by \(\sqrt{x-5}\) to get \(\sqrt{x}+\sqrt{x-4}=\sqrt a\). The left side is strictly increasing and tends to \(\sqrt5+1\) as \(x\to5+\). A second root exists exactly when \(a\gt6+2\sqrt5\approx10.47\). The smallest integer is \(11\).
