Solving quadratic equations 2

Consider a quadratic equation

ax2+bx+c=0ax^2+bx+c=0

where aa, bb, and cc are the coefficients, which we assume are known values. In the last exercise of the previous section we have seen that the solutions of this equation can be calculated using the following formula, which we call the midnight formula (that's not an official name for this formula, but many people and textbooks use it ...):

Theorem 1: midnight formula

The solutions of the quadratic equation

ax2+bx+c=0ax^2+bx+c=0

can be calculated as follows:

x1=−b+b2−4ac2ax_{1}=\frac{-b+ \sqrt{b^2-4ac}}{2a}x2=−b−b2−4ac2ax_{2}=\frac{-b- \sqrt{b^2-4ac}}{2a}

or in short

x1,2=−b±b2−4ac2ax_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

So rather than going through the trouble of completing the square, we simply insert the values of a,ba,b and cc into the midnight formula to calculate the solutions. Of course, the midnight formula is nothing else than actually solving the equation using completing the square (see previous section, last problem).

Let's make some examples.

Example 1: Two solutions
Solution

It is a=3,b=−2a=3, b=-2 and c=−5c=-5. Inserting these values into the midnight formula, we get

x1,2=−(−2)±(−2)2−4⋅3⋅(−5)2⋅3=2±86\begin{array}{lll} x_{1,2}&=&\frac{-(-2)\pm \sqrt{(-2)^2-4\cdot 3\cdot (-5)}}{2\cdot 3}\\ &=&\frac{2\pm 8}{6} \end{array}

So we have x1=106=53‾x_1=\frac{10}{6}=\underline{\frac{5}{3}} and x2=−1‾x_2=\underline{-1}.

Find the solutions of the equation 3x2−2x−5=03x^2-2x-5=0 using the midnight formula.

Example 2: One solution
Solution

It is a=1,b=−2a=1, b=-2 and c=1c=1. Inserting these values into the midnight formula, we get

x1,2=−(−2)±(−2)2−4⋅1⋅12⋅1=2±02=1\begin{array}{lll} x_{1,2} &=& \frac{-(-2)\pm \sqrt{(-2)^2-4\cdot 1\cdot 1}}{2\cdot 1}\\ &=&\frac{2\pm 0}{2}\\&=&1 \end{array}

so x1=x2=1‾x_1=x_2=\underline{1} (only one solution).

Find the solutions of the equation x2−2x+1=0x^2-2x+1=0 using the midnight formula.

Example 3: No solution
Solution

It is a=1,b=1a=1, b=1 and c=1c=1. Inserting these values into the midnight formula, we get

x1,2=−1±12−4⋅1⋅12⋅1=−1±−32\begin{array}{lll} x_{1,2} &=& \frac{-1\pm \sqrt{1^2-4\cdot 1\cdot 1}}{2\cdot 1}\\ &=&\frac{-1\pm \sqrt{-3}}{2} \end{array}

Because the root of a negative number does not exist (if the number set is the real numbers), this equation has no solution.

Find the solutions of the equation x2+x+1=0x^2+x+1=0 using the midnight formula.

Note that the expression under the root of the midnight formula determines the number of solutions of the quadratic equation (see the examples above). If the number under the root is positive, there are two solutions, if the number under the root is 00 we have one solution, and if the number under the root is negative, there is no solution. Let's rephrase this, but first a definition:

Definition 1

Consider the equation

ax2+bx+c=0ax^2+bx+c=0

where the coefficients aa, bb and cc are known numbers. Let's define the value

D=b2−4acD=b^2-4ac

which is called the discriminant (it is the value under the root of the midnight formula).

Theorem 2

Let DD be the discriminant of a quadratic equation. We have

D>0two solutions x1≠x2D=0one solution x1=x2D<0no solutions \begin{array}{lll} D>0 & \text{two solutions } &x_1\neq x_2\\ D=0 & \text{one solution } &x_1=x_2\\ D<0 & \text{no solutions } & \end{array}
Example 4

Without actually solving it, find the number of solutions of the equation

2x2−2+1=02x^2-2+1=0
Solution

We have a=2,b=−2a=2, b=-2 and c=1c=1, thus the discriminant is

D=b2−4ac=(−2)2−4⋅2⋅1=−4<0D=b^2-4ac=(-2)^2-4\cdot 2\cdot 1=-4 <0

so no solution.

And here are some further problems.

Exercise 1

Solve the following quadratic equations using the midnight formula.

  1. 2x2−7x+3=02x^2 - 7x + 3 = 0

  2. 5x+4x2−6=05x +4x^2- 6 = 0

  3. 1+2x−x2=01+2x-x^2 = 0

  4. 5x2=4−8x5x^2 = 4-8x

  5. 5x2−6x−2x+4=05x^2 - 6x -2x + 4 = 0

  6. −x2+70x−1225=0-x^2 + 70x - 1225 = 0

  7. x2−83x+36=0x^2 - 8\sqrt{3} x + 36 = 0

  8. 2x2+x−2=0\sqrt{2}x^2+ x -\sqrt{2} = 0

  9. (1+3)x2+x=−1+3(1 +\sqrt{3})x^2+ x = -1 +\sqrt{3}

  10. 12x2+13x−16=0\frac{1}{2} x^2+\frac{1}{3}x-\frac{1}{6}=0

  11. 75x2+11x−66=−2x275x^2 + 11x - 66 = -2x^2

  12. 0.5x2−2.3x+1.2=00.5x^2 - 2.3x + 1.2 = 0

Solution
  1. x1=0.5‾,x2=3‾x_1=\underline{0.5}, x_2=\underline{3}
  2. x1=−2‾,x2=0.75‾x_1=\underline{-2}, x_2=\underline{0.75}
  3. x1=1−2‾,x2=1+2‾x_1=\underline{1-\sqrt{2}}, x_2=\underline{1+\sqrt{2}}
  4. x1=−2‾,x2=2/5‾x_1=\underline{-2}, x_2=\underline{2/5}
  5. no solution (root is negative)
  6. x1=x2=35‾x_1=x_2=\underline{35} (one solution)
  7. x1=23‾,x2=63‾x_1=\underline{2\sqrt{3}}, x_2=\underline{6\sqrt{3}}
  8. x1=12‾,x2=−2‾x_1=\underline{\frac{1}{\sqrt{2}}}, x_2= \underline{-\sqrt{2}}
  9. x1=1−3‾,x2=0.53−0.5‾x_1=\underline{1-\sqrt{3}}, x_2=\underline{0.5\sqrt{3}-0.5}
  10. x1=−1‾,x2=1/3‾x_1=\underline{-1}, x_2=\underline{1/3}
  11. x1=−1‾,x2=6/7‾x_1=\underline{-1}, x_2=\underline{6/7}
  12. x1=0.6‾,x2=4‾x_1=\underline{0.6}, x_2=\underline{4}
Exercise 2

How many solutions are there? Find out without actually solving the equation.

  1. x2+100x+1=0x^2+100x+1=0

  2. 0.3x2−2.4x+4.8=00.3x^2-2.4x+4.8=0

  3. 16x2+25x+10=016x^2+25x+10=0

  4. x2−3x+cx^2-3x+c (find number of solutions depending on cc).

  5. x2+bx+1=0x^2+bx+1=0 (find number of solutions depending on bb).

Solution

Discriminant D=b2−4acD=b^2-4ac

  1. D=9996>0→D=9996>0 \rightarrow two solutions.
  2. D=0→D=0 \rightarrow one solution.
  3. 16x2+25x+10=016x^2+25x+10=0 D=−15<0→D=-15<0 \rightarrow no solution.
  4. D=9−4cD=9-4c.
    • One solutions if D=9−4c=0→c=9/4=2.25‾D=9-4c=0 \rightarrow c=9/4=\underline{2.25}.
    • No solution if D=9−4c<0→c>2.25‾D=9-4c<0 \rightarrow \underline{c>2.25}.
    • Two solutions if D=9−4c>0→c<2.25‾D=9-4c>0 \rightarrow \underline{c<2.25}.
  5. D=b2−4D=b^2-4
    • One solutions if D=b2−4=0→b=±2‾D=b^2-4=0 \rightarrow b=\underline{\pm 2}.
    • No solution if D=b2−4<0→−2<b<2‾D=b^2-4<0 \rightarrow \underline{-2<b<2}.
    • Two solutions if D=b2−4>0→b<−2 or b>2‾D=b^2-4>0 \rightarrow \underline{b<-2 \text{ or } b>2}.
Exercise 3: Word problems
Solution

  1. xx smaller number. x(x+1)=1122→x1=33‾,x2=−34‾x(x+1)=1122 \rightarrow x_1=\underline{33}, x_2=\underline{-34}.

  2. xx width. x(x+5)=50→x1=5cm‾,x2=−10cmx(x+5)=50 \rightarrow x_1=\underline{5cm}, x_2=-10cm (not possible).

  3. x2+(x+1)2=52→x1=−4x^2+(x+1)^2=5^2 \rightarrow x_1=-4 (not possible) and x2=3‾x_2=\underline{3}, area is 6‾\underline{6}.

  4. xx and yy the two numbers. x+y=9,x2+y2=41→x1=5‾x+y=9, x^2+y^2=41 \rightarrow x_1=\underline{5} and x2=4‾x_2=\underline{4}.

  5. (8+x)(20−x)=0.6⋅8⋅12→x1=16‾(8+x)(20-x)=0.6\cdot 8\cdot 12 \rightarrow x_1=\underline{16} and x2=−4x_2=-4 (not possible).

  6. 15/(x+2)+15/(x−2)=3→3x2−30x−12=0→x1=10.385km/h‾15/(x+2)+15/(x-2)=3 \rightarrow 3x^2-30x-12=0 \rightarrow x_1=\underline{10.385km/h} and x2=−0.385km/hx_2=-0.385 km/h.

  1. The product of two consecutive numbers is 11221122. Find the two numbers (all solutions).
  2. The length of a rectangle is 5cm5cm bigger than its width, and the area is 50cm250cm^2. Find the length and width of the rectangle.
  3. The three sides of a right-angled triangle are xx, x+1x+1, and 55 (longest side). Find the area.
  4. The sum of two numbers is 99, the sum of the squared numbers is 4141. Find the two numbers.
  5. A rectangle has side lengths 8cm and 20cm. Increasing the smaller one by xx and decreasing the larger one by xx, the resulting new rectangle area is 60%60\% of the old rectangle area. Find xx.
  6. A 3 hour river cruise goes 15km upstream and then back again. The river has a current of 2km per hour. What is the speed of the boat on the water?