\(2x(x+2)-5(x+2)=0\). Factor: \((x+2)(2x-5)=0\). Hence \(x=-2\) or \(x=\dfrac52=2.5\). Answer: E.
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2
Task
One correct answer
If \(x+2y-6z=-1\) and \(-y+3z=5\), then \(x=\ldots\)
Solution
From \(-y+3z=5\), \(y-3z=-5\), so \(2y-6z=-10\). Substitute into the first equation: \(x-10=-1\), hence \(x=9\). Answer: A.
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3
Task
One correct answer
Solve the equation \(\dfrac{2}{x}=5\).
Solution
The domain requires \(x\ne0\). Multiply by \(x\): \(2=5x\), so \(x=\dfrac25=0.4\). Answer: D.
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4
Task
Numeric answer
Solve \(||2x-1|-3|=5\). If there is one root, enter it. If there is more than one root, enter the product of all roots.
Solution
Let \(u=|2x-1|\ge0\). Then \(|u-3|=5\), so \(u=8\) or \(u=-2\). Since \(u\ge0\), \(|2x-1|=8\). Thus \(x_1=4.5\) and \(x_2=-3.5\), and their product is \(-15.75\).
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5
Task
One correct answer
Solve the equation \(\dfrac{x}{2}+\dfrac{x}{3}=2\).
Solution
Multiply by \(6\): \(3x+2x=12\). Hence \(5x=12\) and \(x=\dfrac{12}{5}=2.4\). Answer: D.
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6
Task
One correct answer
Solve the equation \(\dfrac{2x-3}{3}=\dfrac{x+1}{6}\).
Solution
Multiply by \(6\): \(2(2x-3)=x+1\). Thus \(4x-6=x+1\), so \(3x=7\) and \(x=\dfrac73\). Answer: E.
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7
Task
One correct answer
Solve the equation \(0.5(3x-4)=\dfrac{x+1}{4}\).
Solution
Multiply by \(4\): \(2(3x-4)=x+1\). Hence \(6x-8=x+1\), so \(5x=9\) and \(x=\dfrac95\). Answer: D.
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8
Task
One correct answer
Find the sum of the roots of \(|x-1|=6\).
Solution
\(|x-1|=6\) gives \(x-1=6\) or \(x-1=-6\). Thus \(x_1=7\), \(x_2=-5\), and their sum is \(2\). Answer: C.
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9
Task
One correct answer
Solve the equation \((x^2+2x)(x-1)=0\).
Solution
\((x^2+2x)(x-1)=x(x+2)(x-1)=0\). Hence \(x=0\), \(x=-2\), or \(x=1\). Answer: C.
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10
Task
One correct answer
Find the product of the roots of \(x^2+6x-55=0\).
Solution
By Vieta's formulas, for \(x^2+px+q=0\) the product of the roots equals \(q\). Here \(q=-55\), so \(x_1x_2=-55\). Answer: A.
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11
Task
One correct answer
Solve the equation \(\dfrac{|x|}{10}=2\).
Solution
Multiply by \(10\): \(|x|=20\). Hence \(x=-20\) or \(x=20\). Answer: B.
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12
Task
One correct answer
Choose the interval containing the solution of \(0.5(x-4)=1.5\).
Solution
Multiply by \(2\): \(x-4=3\), so \(x=7\). The number \(7\) lies in \((4;8]\). Answer: D.
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13
Task
One correct answer
Solve the equation \(|2x-1|=6\).
Solution
We have \(2x-1=6\) or \(2x-1=-6\). Thus \(x=3.5\) or \(x=-2.5\). Answer: B.
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14
Task
One correct answer
Solve the equation \(4x^2=1\).
Solution
Divide by \(4\): \(x^2=\dfrac14\). Hence \(x=\pm\dfrac12\), so \(x=-0.5\) or \(x=0.5\). Answer: E.
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15
Task
One correct answer
If \(x\) and \(y\) satisfy \(2y+4=x\), then \(y=\ldots\)
Solution
From \(2y+4=x\), we get \(2y=x-4\). Therefore \(y=\dfrac{x-4}{2}\). Answer: C.
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16
Task
One correct answer
Solve the equation \(x^2-10=5x+14\).
Solution
Move all terms to one side: \(x^2-5x-24=0\). Factor: \((x-8)(x+3)=0\). Hence \(x=-3\) or \(x=8\). Answer: C.
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17
Task
One correct answer
Find \(x_1^2+x_2^2\), where \(x_1\) and \(x_2\) are the roots of \(x^2-12x+13=0\).
Solution
By Vieta's formulas, \(x_1+x_2=12\) and \(x_1x_2=13\). Thus \(x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=144-26=118\). Answer: E.
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18
Task
One correct answer
Choose the root of \(|x^2-6x|=9\) that belongs to the interval \((-2;1]\).
Solution
There are two cases: \(x^2-6x=9\) or \(x^2-6x=-9\). These give \(x=3\pm3\sqrt2\) and \(x=3\), respectively. Only \(3-3\sqrt2\approx-1.24\) lies in \((-2;1]\). Answer: A.
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19
Task
One correct answer
Which interval contains the solution of \(2x-3=4\)?
Solution
From \(2x-3=4\), we get \(2x=7\), so \(x=3.5\). This lies in \([2;4)\). Answer: D.
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20
Task
One correct answer
Which of the following numbers is a root of \(2|x|=2\)?
Solution
Divide by \(2\): \(|x|=1\). Thus \(x=1\) or \(x=-1\). Of the listed choices, only \(x=-1\) appears. Answer: D.
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21
Task
One correct answer
Solve the equation \( (x+1)(2x-3)=0 \).
Solution
A product is zero when at least one factor is zero: \(x+1=0\) or \(2x-3=0\). Hence \(x=-1\) or \(x=\dfrac32=1.5\). Answer: E.
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22
Task
One correct answer
Solve the equation \(x^2-8x+15=0\).
Solution
Factor: \(x^2-8x+15=(x-3)(x-5)\). Hence \(x=3\) or \(x=5\). Answer: A.
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23
Task
Numeric answer
Solve \(|5-4x|=3\). If the equation has one root, enter it. If it has more than one root, enter the sum of the roots.
Solution
We have \(5-4x=3\) or \(5-4x=-3\). Thus \(x=0.5\) and \(x=2\). Their sum is \(2.5\).
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24
Task
One correct answer
Find the ratio of the smaller root of \(x^2+5x-6=0\) to the larger root.
Solution
\(x^2+5x-6=(x+6)(x-1)\), so the roots are \(-6\) and \(1\). The ratio of the smaller root to the larger is \(-6:1=-6\). Answer: B.
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25
Task
One correct answer
If \(x=t-2\), then \(x^2-t^2=\ldots\)
Solution
Substitute \(x=t-2\): \(x^2-t^2=(t-2)^2-t^2=t^2-4t+4-t^2=4-4t\). Answer: B.
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26
Task
One correct answer
Which of the following numbers is a root of \(\dfrac{5x+8}{3}=1\)?
Solution
Multiply by \(3\): \(5x+8=3\). Then \(5x=-5\), so \(x=-1\). Answer: E.
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27
Task
One correct answer
Solve the equation \(x^2-4x+3=0\).
Solution
Factor: \(x^2-4x+3=(x-1)(x-3)\). Hence \(x=1\) or \(x=3\). Answer: B.
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28
Task
One correct answer
Choose the number that is a root of \(\dfrac{8}{x}=\dfrac25\).
Solution
Cross-multiplying gives \(8\cdot5=2x\). Thus \(40=2x\), so \(x=20\). Answer: A.
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29
Task
One correct answer
Solve the equation \(x^2=25x\).
Solution
Move all terms to one side: \(x^2-25x=0\). Factor: \(x(x-25)=0\). Hence \(x=0\) or \(x=25\). Answer: B.
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30
Task
One correct answer
Solve the equation \(\dfrac{x}{10}=2.5\).
Solution
Multiply by \(10\): \(x=2.5\cdot10=25\). Answer: D.
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31
Task
Numeric answer
Solve \(x+4|x|=3\). If the equation has one root, enter it. If it has more than one root, enter the sum of the roots.
Solution
For \(x\ge0\), \(|x|=x\), so \(5x=3\) and \(x=0.6\). For \(x<0\), \(|x|=-x\), so \(x-4x=3\) and \(x=-1\). The sum of the roots is \(-0.4\).
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32
Task
One correct answer
Choose the root of the equation \(1-5x=0\).
Solution
\(1-5x=0\Rightarrow 5x=1\Rightarrow x=\dfrac15\). Answer: C.
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33
Task
Numeric answer
Solve \(x^4-x^2-20=0\). Enter the product of all its real roots.
Solution
Let \(t=x^2\), where \(t\ge0\). Then \(t^2-t-20=(t-5)(t+4)=0\), so \(t=5\). Hence \(x=\pm\sqrt5\), and the product of the real roots is \(-5\).
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34
Task
One correct answer
Find the sum of the roots of \(x^2+3x-4=0\).
Solution
By Vieta’s formulas, the sum of the roots of \(x^2+3x-4=0\) is \(-3\). Answer: B.
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35
Task
One correct answer
Which of the following intervals contains the root of \(\dfrac{x}{9-x}=\dfrac12\)?
Solution
The domain requires \(x\ne9\). Cross-multiplying gives \(2x=9-x\), so \(3x=9\) and \(x=3\). Thus the root lies in \((2;5]\). Answer: D.
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36
Task
One correct answer
Solve the equation \(2x-3=4\).
Solution
\(2x-3=4\Rightarrow 2x=7\Rightarrow x=3.5\). Answer: B.
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37
Task
One correct answer
Solve the equation \(x(x-4)=x^2+8\).
Solution
Expand: \(x^2-4x=x^2+8\). Cancel \(x^2\): \(-4x=8\), so \(x=-2\). Answer: A.
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38
Task
Numeric answer
Solve \(\bigl||5-5x|-|5x-1|\bigr|=3\) and enter the product of the roots. If there are infinitely many roots, enter 100; if there are no roots, enter 200.
Solution
The breakpoints of the inner absolute values are \(x=0.2\) and \(x=1\). For \(x\in[0.2,1]\), \(|5-5x|-|5x-1|=6-10x\). Thus \(|6-10x|=3\), giving \(x=0.3\) or \(x=0.9\). On the other intervals the outer absolute value equals \(4\), so there are no additional roots. Their product is \(0.27\).
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39
Task
Numeric answer
Solve \(x^3+12x^2+44x+48=0\). Enter the sum of the roots.
Solution
\(x=-2\) is a root. Hence \(x^3+12x^2+44x+48=(x+2)(x^2+10x+24)=(x+2)(x+4)(x+6)\). The roots are \(-2,-4,-6\), whose sum is \(-12\).
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40
Task
Numeric answer
Find the value of the parameter \(a\) for which \((a^2-3a-4)x+16=a^2\) has infinitely many solutions.
Solution
Rewrite the equation as \((a-4)(a+1)x=(a-4)(a+4)\). For infinitely many solutions it must reduce to the identity \(0x=0\). This happens when \(a=4\). For \(a=-1\), it becomes \(0x=-15\), so there is no solution. Therefore \(a=4\).
