Topic 6 Quadratic Formula

6.1 Estimate a Square Root

Can you estimate the irrational numbers \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) and \(\sqrt{7}\) without using a calculator?

Can you estimate the square root \(\sqrt{m^2+n}\), where \(m\) and \(n\) are positive integers?

6.2 Completing the Square

The square root property:
Suppose that \(X^2=d\). Then \(X=\sqrt{d}\) or \(X=-\sqrt{d}\), or simply \(X=\pm\sqrt{d}\).

The square root property provides another method to solve a quadratic equation, completing the square. This method is based on the following observations: \[ x^2+{\mathbf{b}}x+{\mathbf{\left(\frac b2\right)^2}}=\left(x+{\mathbf{\frac b2}}\right)^2, \] and more generally, let \(f(x)=ax^2+bx+c\), and \(h=-\frac{b}{2a}\), then \[ ax^2+bx+c=a(x-h)^2+f(h)=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}. \]

The procedure to rewrite a trinomial as the sum of a perfect square and a constant is called completing the square.

Example 6.1 Solve the equation \(x^2+2x-1=0\).


  1. Isolate the constant. \[x^2+2x=1\]
  2. With \(b=2\), add \(\left(\frac 22\right)^2\) to both sides to complete a square for the binomial \(x^2+bx\). \[ \begin{aligned} x^2+2x+\left(\frac22\right)^2&=1+\left(\frac22\right)^2\\\ \left(x+\frac22\right)^2&=1+1\\ (x+1)^2&=2\\ \end{aligned} \]
  3. Solve the resulting equation using the square root property. \[ \begin{aligned} x+1 =\sqrt2 & \qquad\text{or} & x+1 =-\sqrt2 \\ x =-1+\sqrt2 & \qquad\text{or} & x =-1-\sqrt2 \end{aligned} \]

Note that the solution can also be written as \(x=-1\pm\sqrt2\).

Example 6.2 Solve the equation \(-2x^2+8x-9=0\).


  1. Isolate the constant. \[-2x^2+8x=9\]
  2. Divide by \(-2\) to rewrite the equation in \(x^2+Bx=C\) form \[x^2-4x=-\frac{9}{2}\]
  3. With \(b=-4\), add \(\left(\frac{-4}{2}\right)^2=4\) to both sides to complete the square for the binomial \(x^2-4x\). \[ \begin{aligned} x^2-4x+4&=-\frac{9}{2}+4\\\ (x-2)^2&=-\frac{1}{2} \end{aligned} \]
  4. Solve the resulting equation and simplify. \[ \begin{aligned} x-2 =\frac{\ii}{\sqrt{2}} & \qquad\text{or} & x-2 =-\frac{\ii}{\sqrt{2}} \\ x =2+\frac{\sqrt{2}}{2}\ii & \qquad\text{or} & x =2-\frac{\sqrt{2}}{2}\ii \end{aligned} \]

Another way to complete the square is to use the formula \(ax^2+bx+c=a(x-h)^2+f(h)\), where \(f(h)=ah^2+bh+c\) is the value of the polynomial \(ax^2+bx+c\) at \(x=h\).

6.3 The Quadratic Formula

Using the method of completing the square, we obtain the follow quadratic formula for the quadratic equation \(ax^2+bx+c=0\) with \(a\neq 0\): \[ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}. \]

The quantity \(b^2-4ac\) is called the discriminant of the quadratic equation.

  1. If \(b^2-4ac>0\), the equation has two real solutions.
  2. If \(b^2-4ac=0\), the equation has one real solution.
  3. If \(b^2-4ac<0\), the equation has two imaginary solutions (no real solutions).

Example 6.3 Determine the type and the number of solutions of the equation \((x-1)(x+2)=-3\).


  1. Rewrite the equation in the form \(ax^2+bx+c=0\). \[ \begin{aligned} (x-1)(x+2)&=-3\\ x^2+x+1&=0 \end{aligned} \]
  2. Find the values of \(a\), \(b\) and \(c\). \[ a=1, b=1 ~\text{and}~ c=1. \]
  3. Find the discriminant \(b^2-4ac\). \[ b^2-4ac=1^2-4\cdot 1\cdot 1=-3. \]

The equation has two imaginary solutions.

Example 6.4 Solve the equation \(2x^2-4x+7=0\).


  1. Find the values of \(a\), \(b\) and \(c\). \[ a=2, b=-4 ~\text{and}~ c=7. \]
  2. Find the discriminant \(b^2-4ac\). \[ b^2-4ac=(-4)^2-4\cdot 2\cdot 7=16-56=-40. \]
  3. Apply the quadratic formula and simplify. \[ \begin{aligned} x=&\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\ =&\dfrac{-(-4)\pm\sqrt{-40}}{2\cdot 2}\\ =&\frac{4\pm 2\sqrt{10} \ii}{4}\\ =&1\pm\frac{\sqrt{10}}{2}\ii. \end{aligned} \]

Example 6.5 Find the base and the height of a triangle whose base is three inches more than twice its height and whose area is \(5\) square inches. Round your answer to the nearest tenth of an inch.


  1. We may suppose the height is \(x\) inches. The base can be expressed as \(2x+3\) inches.
  2. By the area formula for a triangle, we have an equation. \[\frac12 x(2x+3)=5.\]
  3. Rewrite the equation in \(ax^2+bx+c=0\) form. \[ \begin{aligned} x(2x+3)&=10\\ 2x^2+3x-10&=0. \end{aligned} \]
  4. By the quadratic formula, we have \[ x=\frac{-3\pm\sqrt{3^2-4\cdot 2\cdot (-10)}}{2\cdot 2}=\frac{-3\pm \sqrt{89}}{4}. \]

Since \(x\) can not be negative, \(x=\frac{-3+\sqrt{89}}{4}\approx 1.6\) and \(2x+3\approx 6.2\). The height and base of the triangle are approximately \(1.6\) inches and 6.2 inches respectively.

6.4 Practice

Problem 6.1 Solve the quadratic equation by the square root property.

  1. \(2x^2-6=0\)
  2. \((x-3)^2=10\)
  3. \(4(x+1)^2+25=0\)

Problem 6.2 Solve the quadratic equation by completing the square.

  1. \(x^2+x-1=0\)
  2. \(x^2+8x+12=0\)
  3. \(3x^2+6x-1=0\)

Problem 6.3 Determine the number and the type of solutions of the given equation.

  1. \(x^2+8x+3=0\)
  2. \(3x^2-2x+4=0\)
  3. \(2x^2-4x+2=0\)

Problem 6.4 Solve using the quadratic formula.

  1. \(x^2+3x-7=0\)
  2. \(2x^2=-4x+5\)
  3. \(2x^2=x-3\)

Problem 6.5 Solve using the quadratic formula.

  1. \((x-1)(x+2)=3\)
  2. \(2x^2-x=(x+2)(x-2)\)
  3. \(\frac12 x^2+x= \frac13\)

Problem 6.6 A triangle whose area is \(7.5\) square meters has a base that is one meter less than triple the height. Find the length of its base and height. Round to the nearest hundredth of a meter.

Problem 6.7 A rectangular garden whose length is \(2\) feet longer than its width has an area 66 square feet. Find the dimensions of the garden, rounded to the nearest hundredth of a foot.