Practice
Solve \(x^2-9x+20=0\).
Find two numbers with sum \(9\) and product \(20\).
\(x^2-9x+20=(x-4)(x-5)\). Hence \(x=4\) or \(x=5\).
Solve \(x^4-10x^2+9=0\).
Set \(u=x^2\), remembering that \(u\ge0\).
Let \(u=x^2\). Then \(u^2-10u+9=0\), so \(u=1\) or \(u=9\).
Thus \(x=\pm1\) or \(x=\pm3\).
Find all pairs \(x,y\) such that \(x+y=8\), \(xy=15\).
Treat \(x,y\) as roots of a quadratic equation.
The numbers \(x,y\) are roots of \(t^2-8t+15=0\).
Thus \((t-3)(t-5)=0\). Therefore \((x,y)=(3,5)\) or \((5,3)\).
Find \(a\) if \(x=3\) is a root of \(x^2-ax+12=0\).
Substitute \(x=3\) into the equation.
We get \(9-3a+12=0\), so \(21-3a=0\).
Therefore \(a=7\).
Let \(x\ne0\) and \(x+\frac1x=5\). Find \(x^2+\frac1{x^2}\).
Square the equality.
\(\left(x+\frac1x\right)^2=x^2+2+\frac1{x^2}=25\).
Hence \(x^2+\frac1{x^2}=23\).
Find \(x,y\) if \(x+y=7\), \(x^2+y^2=25\).
First find \(xy\).
From \(x^2+y^2=(x+y)^2-2xy\), we get \(25=49-2xy\), hence \(xy=12\).
Then \(x,y\) are roots of \(t^2-7t+12=0\), so \(t=3\) or \(t=4\). Answer: \((3,4)\), \((4,3)\).
Solve \(x^4-13x^2+36=0\).
Set \(u=x^2\), then solve the quadratic equation in \(u\).
We get \(u^2-13u+36=0\), so \((u-4)(u-9)=0\).
Since \(u=x^2\), we have \(x^2=4\) or \(x^2=9\). Hence \(x=\pm2,\pm3\).
Solve the system \(x^2+y=12\), \(y^2+x=12\).
Subtract one equation from the other.
Subtracting gives \(x^2-y^2+y-x=0\), so \((x-y)(x+y-1)=0\).
If \(x=y\), then \(x^2+x=12\), so \(x=3\) or \(x=-4\). This gives \((3,3)\), \((-4,-4)\).
If \(x+y=1\), then \(y=1-x\), and \(x^2+1-x=12\). Thus \(x^2-x-11=0\), so \(x=\frac{1\pm3\sqrt5}{2}\), \(y=\frac{1\mp3\sqrt5}{2}\).
Find all integers \(x,y\) such that \(x^2-y^2=21\).
Factor the left side and list factor pairs of \(21\).
\((x-y)(x+y)=21\). The two factors have the same parity, and the product is odd, so both are odd.
The pairs \((1,21)\), \((3,7)\), \((-1,-21)\), \((-3,-7)\) give \((11,10)\), \((5,2)\), \((-11,-10)\), \((-5,-2)\).
For which \(a\) does \(x^2-4x+a=0\) have exactly one real root?
A quadratic equation has exactly one real root when its discriminant is zero.
The discriminant is \(D=16-4a\). We need \(D=0\), so \(16-4a=0\).
Thus \(a=4\).
Find \(x,y\) if \(x+y=6\), \(x^3+y^3=72\).
Express the sum of cubes through \(xy\).
\(x^3+y^3=(x+y)^3-3xy(x+y)=216-18xy\).
From \(216-18xy=72\), we get \(xy=8\). Then \(x,y\) are roots of \(t^2-6t+8=0\), so \((x,y)=(2,4)\), \((4,2)\).
Real numbers \(x,y,z\) satisfy \(x+y+z=6\), \(x^2+y^2+z^2=12\). Prove that \(xy+yz+zx=12\).
Square \(x+y+z=6\).
\((x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\).
Substitute the data: \(36=12+2(xy+yz+zx)\). Hence \(xy+yz+zx=12\).
Let \(x\ne0\) and \(x+\frac1x=3\). Find \(x^3+\frac1{x^3}\).
Use \(u^3\), where \(u=x+\frac1x\).
Let \(u=x+\frac1x=3\). Then \(u^3=x^3+\frac1{x^3}+3\left(x+\frac1x\right)\).
Thus \(27=x^3+\frac1{x^3}+9\), so \(x^3+\frac1{x^3}=18\).
Find all integers \(a\) for which \(x^2-ax+12=0\) has two integer roots.
The roots must be integer divisors of \(12\).
Let the roots be \(m,n\). Then \(mn=12\), \(m+n=a\).
List integer divisor pairs: \((1,12)\), \((2,6)\), \((3,4)\), and the corresponding negative pairs. Thus \(a=13,8,7,-13,-8,-7\).
Solve the system \(x+y=5\), \(x^2-y^2=15\).
Factor \(x^2-y^2\).
\(x^2-y^2=(x-y)(x+y)\). Since \(x+y=5\), we get \(5(x-y)=15\), so \(x-y=3\).
Solving \(x+y=5\), \(x-y=3\), gives \(x=4\), \(y=1\).
Solve \( (x^2-3x)^2-2(x^2-3x)-8=0 \).
Set \(u=x^2-3x\).
Let \(u=x^2-3x\). Then \(u^2-2u-8=0\), so \(u=4\) or \(u=-2\).
If \(x^2-3x=4\), then \(x^2-3x-4=0\), so \(x=4\) or \(x=-1\). If \(x^2-3x=-2\), then \(x^2-3x+2=0\), so \(x=1\) or \(x=2\).
Real numbers \(x,y,z\) satisfy \(x+y+z=3\), \(x^2+y^2+z^2=3\). Prove that \(x=y=z=1\).
Consider \((x-1)^2+(y-1)^2+(z-1)^2\).
\((x-1)^2+(y-1)^2+(z-1)^2=x^2+y^2+z^2-2(x+y+z)+3=3-6+3=0\).
A sum of squares is zero only when all squares are zero. Hence \(x=y=z=1\).
Solve the system \(x^2+2y=9\), \(y^2+2x=9\).
Subtract the equations and factor the result.
Subtract: \(x^2-y^2+2y-2x=0\), so \((x-y)(x+y-2)=0\).
If \(x=y\), then \(x^2+2x=9\), so \(x=-1\pm\sqrt{10}\). This gives two pairs \((-1+\sqrt{10},-1+\sqrt{10})\), \((-1-\sqrt{10},-1-\sqrt{10})\).
If \(x+y=2\), then \(y=2-x\). Substitution gives \(x^2+4-2x=9\), so \(x^2-2x-5=0\). Hence \(x=1\pm\sqrt6\), \(y=1\mp\sqrt6\).
Find all positive integers \(x,y\) such that \(xy=x+y+5\).
Move everything to the left and add \(1\) to get a product.
We have \(xy-x-y=5\). Add \(1\): \((x-1)(y-1)=6\).
Since \(x,y\) are positive integers, \(x-1\) and \(y-1\) are positive divisors of \(6\). The pairs \((1,6)\), \((2,3)\), \((3,2)\), \((6,1)\) give \((x,y)=(2,7),(3,4),(4,3),(7,2)\).
Find all real \(x,y\) if \(x+y=2\) and \(x^4+y^4=2\).
First express everything through \(p=xy\).
Let \(p=xy\). Then \(x^2+y^2=4-2p\).
\(x^4+y^4=(x^2+y^2)^2-2x^2y^2=(4-2p)^2-2p^2=16-16p+2p^2\).
Since this equals \(2\), \(16-16p+2p^2=2\), so \(p^2-8p+7=0\). Thus \(p=1\) or \(p=7\). But real \(x,y\) with \(x+y=2\) require \(p\le1\). Hence \(p=1\), and \(x,y\) are roots of \(t^2-2t+1=0\). Therefore \(x=y=1\).
Find all integers \(a\) for which \(x^2-(a+1)x+a+6=0\) has root \(x=3\) or \(x=4\).
Substitute \(x=3\) and \(x=4\) separately.
For \(x=3\): \(9-3(a+1)+a+6=0\), so \(12-2a=0\), hence \(a=6\).
For \(x=4\): \(16-4(a+1)+a+6=0\), so \(18-3a=0\), hence \(a=6\). Thus the only value is \(a=6\). For this value, both \(3\) and \(4\) are indeed roots.
Real numbers \(x,y\) satisfy \(x^2+y^2=2x+4y-5\). Prove that \(x=1\), \(y=2\).
Move everything to the left and complete squares.
Rewrite as \(x^2-2x+y^2-4y+5=0\).
This is \((x-1)^2+(y-2)^2=0\). A sum of squares is zero only when \(x=1\), \(y=2\).
Find all real triples \(x,y,z\) such that \(x+y+z=3\), \(xy+yz+zx=3\), \(xyz=1\).
Treat \(x,y,z\) as roots of a cubic polynomial.
The numbers \(x,y,z\) are roots of \(t^3-3t^2+3t-1\).
But \(t^3-3t^2+3t-1=(t-1)^3\). Therefore all three roots equal \(1\), so \(x=y=z=1\).
Find all positive integers \(x,y,z\) such that \(xyz=x+y+z+2\).
Assume \(x\le y\le z\) and bound \(x\).
Without loss of generality, let \(x\le y\le z\). If \(x\ge3\), then \(xyz\ge3yz\), while the right side is at most \(3z+2\); for \(y\ge3\), \(z\ge3\), this is impossible. Hence \(x=1\) or \(x=2\).
If \(x=1\), then \(yz=y+z+3\), so \((y-1)(z-1)=4\). This gives \((y,z)=(2,5),(3,3),(5,2)\).
If \(x=2\), then \(2yz=y+z+4\). For \(y=1\), \(z=5\); for \(y=2\), \(z=2\); for \(y\ge3\), the left side is too large: \(2yz-y-z-4\ge6z-3-z-4=5z-7>0\). Taking permutations into account, all solutions are permutations of \((1,2,5)\), \((1,3,3)\), and \((2,2,2)\).
Solve the system \(x+y+xy=11\), \(x^2+y^2=25\).
Set \(s=x+y\), \(p=xy\).
Let \(s=x+y\), \(p=xy\). Then \(s+p=11\), and \(x^2+y^2=s^2-2p=25\).
Since \(p=11-s\), we get \(s^2-2(11-s)=25\), so \(s^2+2s-47=0\). Thus \(s=-1\pm4\sqrt3\), \(p=11-s\).
Real \(x,y\) require \(s^2-4p\ge0\), which holds for both values. Then \(x,y\) are roots of \(t^2-st+p=0\). Thus for each \(s=-1\pm4\sqrt3\), \(p=11-s\), the pairs \((x,y)\) are the two permutations of the roots of \(t^2-st+p=0\).
Positive numbers \(x,y,z\) have the following property: the values \(x+2y^2+2z^2\), \(y+2z^2+2x^2\), \(z+2x^2+2y^2\) are equal. Must \(x=y=z\)?
Subtract two equal expressions. A factor \(x-y\) should appear.
Answer: no, not necessarily. We show how to find a counterexample.
From the equality of the first two expressions, \(x+2y^2=y+2x^2\). Move all terms to one side:
\[x-y-2(x^2-y^2)=0.\]
Thus \((x-y)(1-2(x+y))=0\). Therefore for each pair of variables, either they are equal or their sum is \(\frac12\).
Take \(x=y=\frac16\), \(z=\frac13\). Then \(x+y=\frac13\), but \(x+z=y+z=\frac12\). Check:
\[x+2y^2+2z^2=\frac16+\frac1{18}+\frac2{9}=\frac49,\]
and the other two expressions are also equal to \(\frac49\). However, \(x,y,z\) are not all equal. Hence the statement is false.