## 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.

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 $$-2x^2+8x-9=0$$.

Solution.

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 $$x=h$$.

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 $$(x-1)(x+2)=-3$$.

Solution.

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 $$2x^2-4x+7=0$$.

Solution.

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.

Solution.

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.