Finding the Value that Completes the Square

Finding the Value that Completes the Square exercise

• Explain how to solve the given function.

  Hints

  The standard form of a quadratic equation is given by $ax^2+bx+c=0$.

  Keep the following in mind:

  $(a+b)^2=a^2+2ab-b^2$ or $(a-b)^2=a^2-2ab-b^2$.

  $x^2+bx+c=0$ is equivalent to $x^2+bx+\left(\frac b2\right)^2=-c+\left(\frac b2\right)^2$.

  Factoring, we get $x^2+bx+\left(\frac b2\right)^2=\left(x+\frac b2\right)^2$.

  Solution

  How should one solve such an equation?

  1. First rewrite the equation so that the coefficient of the highest power is equal to $1$: we have to multiply by $-45$. This leads to $x^2-80x+700=0$.
  2. Next move the constant to the right using opposite operations: $x^2-80x=-700$.
  3. Now find the $c$ value to form a perfect square trinomial, namely $c=\left(\frac{-80}2\right)^2=40^2=1600$.
  4. Add this value on both sides of the equation and combine the terms on the right-hand side: $x^2-80x+1600=-700+1600=900$.
  5. Factor the trinomial on the left-hand side: $(x-40)^2=900$.
  6. Almost done: take the square root on both sides to get $x-40=\pm 30$.
  7. Last but not least you get the solutions by adding $40$ on both sides: $x=30+40=70$ or $x=-30+40=10$.

• Solve the given quadratic equation.

  Hints

  Let's have a look at $x^2+4x-5=0$:

  1. Add $5$ to $x^2+4x=5$.
  2. $c=\left(\frac42\right)^2=4$
  3. Add $c=4$ to $x^2+4x+4=9$.

  Looking at $(a+b)^2=a^2+2ab+b^2$ and the trinomial $x^2+4x+4$, we can conclude that $a=x$ and $b=2$. We then have that

  $(x+2)^2=x^2+2x(2)+2^2=x^2+4x+4$.

  Now we can proceed as follows: $x^2+4x+4=9$ is equivalent to

  $(x+2)^2=9$.

  Now we take the square root to get $x+2=\pm 3$.

  Last we subtract $2$ to get the solutions

  $x=3-2=1$

  or

  $x=-3-2=-5$.

  Solution

  Let's practice finding the solution of a quadratic equation by finding the value that completes the square:

  1. Divide the equation on both sides by $-0.01$ to get the coefficient of the highest power equal to $1$: $x^2-100x+900=0$.
  2. Move the constant to the right: $x^2-100x=-900$.
  3. Now we have to do the main part of the work: find the $c$ value to form a perfect square trinomial, namely $c=\left(\frac{-100}2\right)^2=50^2=2500$.
  4. Add this value to both sides of the equation: $x^2-100x+2500=-900+2500=1600$.
  5. Factor the trinomial: $(x-50)^2=1600$.
  6. Take the square root: $x-50=\pm 40$.
  7. Add $50$ to get the solutions $x=40+50=90$ or $x=-40+50=10$.

• Find the missing value $c$ that completes the square.

  Hints

  Take a look at a quadratic binomial

  $x^2+2bx$.

  The missing value completes this term to

  $x^2+2bx+b^2-b^2=(x+b)^2-b^2$.

  In general the missing term is given as the square of the coefficient of the linear term divided by $2$.

  Let's have a look at another example, $x^2+6x$.

  The missing term is $\left(\frac62\right)^2=3^2=9$.

  Solution

  For $x^2+bx$, the missing value $c$ is given by $c=\left(\frac b2\right)^2$.

  Adding this value to the binomial above we get a perfect square trinomial which we can factor out.

  1. $x^2+12x$ - here we have $b=12$. Squaring half of it gives us $c=\left(\frac{12}2\right)^2=6^2=36$.
  2. $x^2-8x$ - we can conclude with $b=12$ that $c=\left(\frac{-8}2\right)^2=4^2=16$.
  3. $x^2+16x$ - with $b=12$ we get $c=\left(\frac{16}2\right)^2=8^2=64$.
  4. $x^2+18x$ - here we have $b=18$. Thus we get $c=\left(\frac{18}2\right)^2=9^2=81$.

• Decide which equation(s) will give General Good's jetpack the right distance to jump.

  Hints

  All the trinomials on the left-hand sides are given in perfect square form.

  For example, we have $x^2+140x+4900=(x+70)^2$.

  After factoring, take the square of both sides.

  Let's have a look at an example:

  $(x+4)^2=16$.

  This leads to $x+4=\pm 4$.

  Subtracting $4$ on both sides leads to $x=-8$ or $x=0$.

  Here we have the distance $|-8-0|=8$.

  Solution

  We solve each equation above and subtract the resulting solutions to check if the distance is $80$.

  Each trinomial is in a perfect square form:

  1. $x^2-120x+3600=1600$ is equivalent to $(x-60)^2=1600$. Taking the square root gives us $x-60=\pm 40$ and thus $x=40+60=100$ or $x=-40+60=20$. The difference of both solutions is $100-20=80$ $~~~~$✓
  2. $x^2-120x+3600=900$ is equivalent to $(x-60)^2=1600$. Again taking the square root leads to $x-60=\pm 30$ and thus $x=30+60=90$ or $x=-30+60=30$. The difference of both solutions is $90-30=60$. This equation would not help her get from tree to tree.
  3. $x^2+80x+1600=2500$ - we rewrite this as $(x+40)^2=2500$ and take the square root to $x+40=\pm 50$. The resulting solutions are $x=50-40=10$ or $x=-50-40=-90$. The difference of both solutions is $10-(-90)=100$. This equation would not help her get from tree to tree.
  4. $x^2+80x+1600=1600$ - the difference to the equation above is the right side: $(x+40)^2=1600$. Taking the square root results in $x+40=\pm 40$ and the solutions are $x=40-40=0$ or $x=-40-40=-80$. The difference of both solutions is $0-(-80)=80$ $~~~~$✓
  5. $x^2+140x+4900=1600$ - factoring leads to $(x+70)^2=1600$. Now we take the square root to $x+70=\pm 40$ to get the solutions $x=40-70=-30$ or $x=-40-70=-110$. The difference of those solutions is $-30-(-110)=80$ $~~~~$✓
  6. $x^2-140x+4900=2500$ leads to $(x-70)^2=4900$ and, by taking the square root, to $x-70=\pm70$. This gives us the solutions $x=70-70=0$ or $x=-70-70=-140$ with the difference $0-(-140)=140$. This equation would not help her get from tree to tree.

• Identify the polynomial which is equal to $(a+b)^2$.

  Hints

  Factor out the squared binomial $(a+b)^2=(a+b)(a+b)$ using the FOIL method.

  For example, $x^2+6x+9=(x+3)^2$.

  Solution

  Finding the value that completes the square leads to the following:

  $(a+b)^2=a^2+2ab+b^2$.

  So if you already have a binomial like $x^2+6x$ you have to find a value that completes this binomial to a perfect trinomial which can be factored.

  This value for $x^2+6x$ is $\left(\frac62\right)^2=3^2=9$, giving us that

  $x^2+6x+9-9=(x+3)^2-9$.

  The trinomial on the left $x^2+6x+9$ is the desired perfect trinomial.

• Find the two solutions of the following equation.

  Hints

  You have to divide both sides of the equation by the coefficient of the highest power, unless it is already $1$.

  The value $c$ completing the square of $x^2+bx$ is given by $c=\left(\frac b2\right)^2$.

  You can check your solutions by putting them in the equation above.

  Solution

  Step by step we solve the equation:

  1. Divide by $4$ to get a coefficient of the highest power equal to $1$: $x^2+8x-20=0$.
  2. Move the constant to the right: $x^2+8x=20$.
  3. The value that completes the square on the left to a perfect square trinomial is given by $c=\left(\frac82\right)^2=4^2=16$.
  4. Add $c=16$: $x^2+8x+16=20+16=36$.
  5. Factor the trinomial: $(x+4)^2=36$.
  6. Take the square root: $x+4=\pm 6$.
  7. Last but not least, you get the solutions by subtracting $4$ on both sides: $x=6-4=2$ or $x=-6-4=-10$.