# 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 \(x^2+2x-1=0\).

*Solution*.

- Isolate the constant. \[x^2+2x=1\]
- With \(b=2\), add \(\left(\frac 22\right)^2\) to both sides to complete a square for the binomial \(x^2+bx\). \[ \begin{aligned} x^2+2x+\left(\frac22\right)^2&=1+\left(\frac22\right)^2\\\ \left(x+\frac22\right)^2&=1+1\\ (x+1)^2&=2\\ \end{aligned} \]
- 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 \(-2x^2+8x-9=0\).

*Solution*.

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

- If \(b^2-4ac>0\), the equation has two real solutions.
- If \(b^2-4ac=0\), the equation has one real solution.
- 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 \((x-1)(x+2)=-3\).

*Solution*.

- Rewrite the equation in the form \(ax^2+bx+c=0\). \[ \begin{aligned} (x-1)(x+2)&=-3\\ x^2+x+1&=0 \end{aligned} \]
- Find the values of \(a\), \(b\) and \(c\). \[ a=1, b=1 ~\text{and}~ c=1. \]
- 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 \(2x^2-4x+7=0\).

*Solution*.

- Find the values of \(a\), \(b\) and \(c\). \[ a=2, b=-4 ~\text{and}~ c=7. \]
- Find the discriminant \(b^2-4ac\). \[ b^2-4ac=(-4)^2-4\cdot 2\cdot 7=16-56=-40. \]
- 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.

*Solution*.

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

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

**Problem 6.2 **Solve the quadratic equation by completing the square.

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

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

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

**Problem 6.4 **Solve using the quadratic formula.

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

**Problem 6.5 **Solve using the quadratic formula.

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

**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.