Exercises — Quadratic equations — factoring, completing the square

A quadratic must have an $x^2$ term with a non-zero coefficient (that is the meaning of $a\ne 0$), and no power of $x$ higher than $2$, and no $x$ in a denominator.

• (a) Quadratic. $a=3,b=-7,c=2$. (Read the coefficients straight off.)
• (b) Not quadratic — highest power is $x^1$. This is a linear equation.
• (c) Quadratic. Rewrite as $x^2-9=0$, so $a=1,b=0,c=-9$. A missing $x$ term just means $b=0$.
• (d) Not quadratic — the $\frac{1}{x}$ term is $x^{-1}$, a negative power. Not a polynomial.

The Zero Product Property says: if a product is $0$, one factor is $0$. So $x-2=0$ or $x-3=0$. Solving each tiny linear equation: $x-2=0\Rightarrow x=2,x-3=0\Rightarrow x=3.$ Solutions: $x=2$ and $x=3$. Notice the sign flips: the factor holds $-2$ but the root is $+2$.

This is monic ($a=1$), so we hunt two numbers that multiply to $c=12$ and add to $b=7$. Factor pairs of $12$: $(1,12),(2,6),(3,4)$. Which sum to $7$? $3+4=7$. ✓ So $x^2+7x+12=(x+3)(x+4)=0$. Zero Product Property: $x+3=0$ or $x+4=0$, giving $x=-3\text{or}x=-4.$

Step 1 — Isolate constant. Move $c$ to the right: $x^2+8x=-7$. Step 2 — Half the middle, then square. Take half of $b=8$: that is $4$; square it: $4^2=16$. (This $16$ is exactly the constant a perfect square $(x+4{)}^2$ needs, since $(x+4{)}^2=x^2+8x+16$.) Step 3 — Add to BOTH sides (balance the equation): $x^2+8x+16=-7+16=9.$ Step 4 — Fold into a square: $(x+4{)}^2=9$. Step 5 — Undo the square by taking $\pm \sqrt{}$ (both roots, because $(+3{)}^2$ and $(-3{)}^2$ each give $9$): $x+4=\pm 3.$ Step 6 — Solve: $x=-4+3=-1$ or $x=-4-3=-7$.

No middle term, so completing the square is overkill — just isolate $x^2$. $3x^2=27\Rightarrow x^2=9\Rightarrow x=\pm 3.$ Both signs matter: $(-3{)}^2=9$ too. Solutions: $x=3$ and $x=-3$.

Here $a=6,b=11,c=-10$. Because $a\ne 1$ we cannot just "multiply to $c$"; we use $ac$. Step 1 — Target product: $ac=6\times (-10)=-60$. Step 2 — Two numbers multiplying to $-60$, adding to $11$. Since the product is negative, one is positive and one negative. Try $15$ and $-4$: $15\times (-4)=-60$ ✓, $15+(-4)=11$ ✓. Step 3 — Split the middle term $11x$ into $15x-4x$: $6x^2+15x-4x-10=0.$ Step 4 — Group into two pairs: $(6x^2+15x)+(-4x-10)=0.$ Step 5 — Pull out the GCF of each pair: $3x(2x+5)-2(2x+5)=0.$ (Notice both brackets became $(2x+5)$ — that's the sign the method is working.) Step 6 — Factor the shared bracket: $(2x+5)(3x-2)=0$. Step 7 — Solve each factor: $2x+5=0\Rightarrow x=-\frac{5}{2}$, and $3x-2=0\Rightarrow x=\frac{2}{3}$.

Choosing: $ac=2\times (-3)=-6$. Two integers multiplying to $-6$ and adding to $-5$? $-6$ and $+1$: $(-6)(1)=-6$ ✓, $-6+1=-5$ ✓. Clean integers exist, so factoring (AC method) is faster than completing the square. Split: $2x^2-6x+x-3=0$. Group: $(2x^2-6x)+(x-3)=0\Rightarrow 2x(x-3)+1(x-3)=0$. Factor: $(x-3)(2x+1)=0$. Solve: $x-3=0\Rightarrow x=3$; $2x+1=0\Rightarrow x=-\frac{1}{2}$. Solutions: $x=3,x=-\frac{1}{2}$.

Set $h=0$: $-5t^2+20t=0$. Spot the common factor $-5t$ (both terms share $-5$ and $t$): $-5t(t-4)=0.$ Zero Product Property: $-5t=0\Rightarrow t=0$, or $t-4=0\Rightarrow t=4$. Meaning: $t=0$ is the launch (the ball starts on the ground); $t=4\text{ s}$ is when it lands again. Both are genuine roots — a projectile touches height zero twice, matching the two red dots on our opening parabola.

Step 1 — Factor $a=2$ out of the $x$-terms (a perfect square needs the $x^2$ coefficient to be $1$ inside): $2\left(x^2-6x\right)+3=0.$ Step 2 — Half of $-6$ is $-3$; square it: $9$. Add and subtract $9$ inside the bracket: $2\left(x^2-6x+9-9\right)+3=0.$ Step 3 — Fold the perfect square, keep the leftover $-9$: $2\left[(x-3{)}^2-9\right]+3=0=2(x-3{)}^2-18+3=2(x-3{)}^2-15=0.$ Vertex form: $2(x-3{)}^2=15$. Comparing to $a(x-h{)}^2=k$: $h=3$. Solve: $(x-3{)}^2=\frac{15}{2}$, so $x-3=\pm \sqrt{\frac{15}{2}}=\pm \frac{\sqrt{30}}{2}$, giving $x=3\pm \frac{\sqrt{30}}{2}.$ Vertex: the lowest point sits at $x=h=3$; its height is $y=2(0)-15=-15$, so the vertex is $(3,-15)$. See 2.3.4-Parabolas-and-vertex-form for why $(h,k)$ read straight off this form.

"Exactly one solution" means the parabola touches the ground at a single point — the two red dots have merged into one. Completing the square exposes this cleanly. $x^2-6x=-c$. Half of $-6$ is $-3$, squared is $9$. Add $9$ to both sides: $(x-3{)}^2=9-c.$ A square equals a single value only when the right side is $0$ (then $x-3=0$ is the only option; a positive right side gives two roots, a negative gives none). So $9-c=0\Rightarrow \boxed{c=9}$, and the single solution is $x=3$. (Cross-check with 2.1.17-Quadratic-formuland-discriminant: the discriminant $b^2-4ac=36-4c=0\Rightarrow c=9$. Same answer.)

Complete the square: $x^2+2x=-5$. Half of $2$ is $1$, squared is $1$. Add $1$: $(x+1{)}^2=-5+1=-4.$ Now the crunch: a real number squared is never negative, yet the right side is $-4$. So no real solution exists — the parabola $y=x^2+2x+5$ floats entirely above the ground and never touches $y=0$ (its vertex is at $(-1,4)$, above the axis). If we allow imaginary numbers, $x+1=\pm \sqrt{-4}=\pm 2i$, giving $x=-1\pm 2i$ — but over the real line, zero roots.

A perfect square $(x+m{)}^2=x^2+2mx+m^2$. Match term by term against $x^2+kx+9$:

• constant: $m^2=9\Rightarrow m=3$ or $m=-3$.
• middle: $k=2m$. So $m=3\Rightarrow k=6$ giving $(x+3{)}^2=0\Rightarrow x=-3$ (repeated). And $m=-3\Rightarrow k=-6$ giving $(x-3{)}^2=0\Rightarrow x=3$ (repeated). Answer: $k=6$ (root $x=-3$) or $k=-6$ (root $x=3$). Both make the discriminant $k^2-36=0$, confirming the single-touch case.

The Zero Product Property lets us solve $(x-p)(x-q)=0$ by setting each factor to zero separately. When we square-root both sides we must write ::: $\pm$ (both signs) A negative number on the right after completing the square means ::: no real roots (the parabola misses the x-axis)