Linear and quadratic equations and inequalities

Unit 5 Linear and quadratic equations and inequalities

Key unit competence: Solve daily life problems algebraically or graphically using linear and quadratic equations or inequalities.

Key ideas

• $|x-a|<r⟺a-r<x<a+r$; $|x-a|>r⟺x<a-r$ or $x>a+r$ ($r>0$).
• Parametric equations such as $mx+2=x+m$: discuss by cases on the parameter (e.g. $m=1$ gives infinitely many or no solutions).
• Quadratic $a{x}^2+bx+c=0$: $\Delta =b^2-4ac$; roots $\frac{-b\pm \sqrt{\Delta }}{2a}$; $S=x_1+x_2=-\frac{b}{a}$, $P=x_1{x}_2=\frac{c}{a}$. Two numbers with sum $S$ and product $P$ are roots of $x^2-Sx+P=0$.
• Sign of $a{x}^2+bx+c$: same sign as $a$ outside the roots, opposite sign between the roots. If $\Delta <0$ it always has the sign of $a$.
• Equations reducible to quadratics: biquadratic $a{x}^4+b{x}^2+c=0$ (put $y=x^2$); irrational $\sqrt{f(x)}=g(x)$ (square both sides, then check — reject extra roots, require $g(x)\ge 0$); reciprocal equations (divide by $x^2$ and set $y=x+\frac{1}{x}$).
• Simultaneous quadratic equations: substitute the linear equation into the quadratic one.

Formulas to know

Common misconceptions

• "When solving $\sqrt{x+2}=x$, every solution of $x+2=x^2$ is valid." No: $x=-1$ gives $\sqrt{1}=1\ne -1$; always check.
• "$x^2>4$ means $x>2$." No: $x<-2$ or $x>2$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Solve $|x+2|\le 5$ and represent it as an interval.

   Show answer

   $-7\le x\le 3$: $[-7,3]$.

   Common mistake: Writing $x+2\le \pm 5$.

2. A ball is thrown so that $h=-5t^2+20t$. When is it higher than 15 m?

   Show answer

   $-5t^2+20t>15⟺t^2-4t+3<0⟺1<t<3$ seconds.

   Common mistake: Forgetting to reverse the sign when dividing by $-5$.

3. Solve $x^2-2x-15>0$ using a sign table.

   Show answer

   Roots $-3$ and 5. Positive outside: $x\in ]-\infty ,-3[\cup ]5,+\infty [$.

   Common mistake: Taking the region between the roots.

4. Solve the system $x+y=7$, $x^2+y^2=29$.

   Show answer

   $y=7-x$: $2x^2-14x+20=0$, $x^2-7x+10=0$: $(2,5)$ or $(5,2)$.

   Common mistake: Writing $(x+y{)}^2=x^2+y^2$.

5. Solve the irrational equation $\sqrt{3x+1}=x-3$.

   Show answer

   $3x+1=x^2-6x+9$, $x^2-9x+8=0$, $x=1$ or $x=8$. Check: $x=1$ gives $2\ne -2$ (reject); $x=8$: $5=5$. Solution $x=8$.

   Common mistake: Keeping $x=1$.

6. Find $m$ so that the roots of $x^2-6x+m=0$ satisfy $x_1=2x_2$.

   Show answer

   $x_1+x_2=3x_2=6$, $x_2=2$, $x_1=4$; $m=x_1{x}_2=8$.

   Common mistake: Using $x_1{x}_2=-6$.

7. Solve the biquadratic equation $4x^4-17x^2+4=0$.

   Show answer

   $y=x^2$: $4y^2-17y+4=0$, $y=4$ or $y=\frac{1}{4}$; $x=\pm 2$ or $x=\pm \frac{1}{2}$.

   Common mistake: Forgetting the negative square roots.

8. Solve the reciprocal equation $2x^4-5x^3+6x^2-5x+2=0$ by dividing by $x^2$ and putting $y=x+\frac{1}{x}$.

   Show answer

   Dividing by $x^2$: $2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+6=0$, and $x^2+\frac{1}{x^2}=y^2-2$: $2y^2-4-5y+6=0$, $2y^2-5y+2=0$, $y=2$ or $y=\frac{1}{2}$. $x+\frac{1}{x}=2\Rightarrow x=1$; $x+\frac{1}{x}=\frac{1}{2}$ has no real roots. Solution $x=1$.

   Common mistake: Writing $x^2+\frac{1}{x^2}=y^2$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. What is the sign of $x^2+2x+5$?	a. $x=6$
   2. Solve the system $x+y=7$, $xy=12$.	b. Always positive
   3. Solve $\frac{x-1}{x+2}\le 0$.	c. $-2<x\le 1$
   4. Solve $\sqrt{x+3}=x-3$.	d. $(3,4)$ and $(4,3)$

   Show answer

   1 → b (Always positive) · 2 → d ($(3,4)$ and $(4,3)$) · 3 → c ($-2<x\le 1$) · 4 → a ($x=6$)

   • Step 1. $\Delta =4-20=-16<0$: no real roots.
     Step 2. It keeps the sign of $a=1$: always positive.
   • Step 1. $x$ and $y$ are the roots of $t^2-7t+12=0$.
     Step 2. $t=3$ or $t=4$: the pairs $(3,4)$ and $(4,3)$.
   • Step 1. Numerator zero at 1; denominator zero at $-2$ (always excluded).
     Step 2. Sign table: negative between $-2$ and 1.
   • Step 1. Condition $x-3\ge 0$. Square: $x+3=x^2-6x+9$.
     Step 2. $x^2-7x+6=0$: $x=1$ or $x=6$.
     Step 3. $x=1$ fails the condition. Check $x=6$: $\sqrt{9}=3$. ✓

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   5 m and 12 m  ·  $x\ge 3$  ·  $-2\le x\le 3$  ·  $x=\pm 2$ or $x=\pm 3$
   1. A rectangle has perimeter 34 m and diagonal 13 m. Its sides are .
   2. Solve $-x^2+x+6\ge 0$. →
   3. Solve $x^4-13x^2+36=0$. →
   Show answer

   1: 5 m and 12 m · 2: $-2\le x\le 3$ · 3: $x=\pm 2$ or $x=\pm 3$

   • Step 1. $x+y=17$ and $x^2+y^2=169$.
     Step 2. $(x+y{)}^2=x^2+y^2+2xy$: $289=169+2xy$, so $xy=60$.
     Step 3. Roots of $t^2-17t+60=0$: 5 and 12.
   • Step 1. Multiply by $-1$ and reverse: $x^2-x-6\le 0$.
     Step 2. Roots $-2$ and 3; the expression is $\le 0$ between them.
   • Step 1. Put $y=x^2$: $y^2-13y+36=0$, so $y=4$ or $y=9$.
     Step 2. $x^2=4$ gives $\pm 2$; $x^2=9$ gives $\pm 3$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. For which $k$ does $x^2+kx+9=0$ have a double root?	a. $x=5$ or $x=-1$
   2. Solve $|2x-1|=x+4$.	b. If $m\ne 2$: $x=3$; if $m=2$: every real $x$
   3. Find $m$ so that $x=2$ is a root of $x^2-2mx+m+6=0$.	c. $m=\frac{10}{3}$
   4. Discuss $(m-2)x=3m-6$ according to $m$.	d. $k=6$ or $k=-6$

   Show answer

   1 → d ($k=6$ or $k=-6$) · 2 → a ($x=5$ or $x=-1$) · 3 → c ($m=\frac{10}{3}$) · 4 → b (If $m\ne 2$: $x=3$; if $m=2$: every real $x$)

   • Step 1. A double root needs $\Delta =0$: $k^2-36=0$.
     Step 2. $k=\pm 6$.
   • Step 1. Need $x+4\ge 0$.
     Step 2. $2x-1=x+4$ gives $x=5$.
     Step 3. $2x-1=-(x+4)$ gives $3x=-3$, $x=-1$. Both satisfy $x+4\ge 0$.
   • Step 1. Substitute $x=2$: $4-4m+m+6=0$.
     Step 2. $10-3m=0$, so $m=\frac{10}{3}$.
   • Step 1. $3m-6=3(m-2)$.
     Step 2. If $m\ne 2$, divide: $x=3$.
     Step 3. If $m=2$: $0x=0$, true for every $x$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $-1<m<3$  ·  $x=4$  ·  $x=\frac{1}{2}$ or $x=3$  ·  $m>1$
   1. Solve $\sqrt{x+5}=x-1$. →
   2. Solve $2x^2-7x+3=0$. →
   3. For which $m$ is $x^2+(m-1)x+1>0$ for every real $x$? →
   Show answer

   1: $x=4$ · 2: $x=\frac{1}{2}$ or $x=3$ · 3: $-1<m<3$

   • Step 1. Condition $x\ge 1$. Square: $x+5=x^2-2x+1$.
     Step 2. $x^2-3x-4=(x-4)(x+1)=0$.
     Step 3. Reject $x=-1$. Check $x=4$: $\sqrt{9}=3$. ✓
   • Step 1. $\Delta =49-24=25$.
     Step 2. $x=\frac{7\pm 5}{4}$: $x=3$ or $x=\frac{1}{2}$.
   • Step 1. Since $a=1>0$, we need $\Delta <0$.
     Step 2. $(m-1{)}^2-4<0$, so $-2<m-1<2$.
     Step 3. $-1<m<3$.

Take the Unit 5 quiz → · Back to contents