Solve the quadraic equation:
$$2x^2-18x-27=0 $$
Quick Answer
Since the discriminant $$\Delta >0$$, the quadratic equation has two distinct real roots.
$$ \Delta=540$$
$$\begin{cases} x_1=\dfrac{9}{2}+\dfrac{3}{2}\sqrt{15} \\ x_2=\dfrac{9}{2}-\dfrac{3}{2}\sqrt{15} \end{cases}$$
In decimal notation,
$$\begin{cases} x_1=10.309475019311 \\ x_2=-1.3094750193111 \end{cases}$$
Detailed Steps on Solution
Solve the quadratic equation: $$2x² - 18x - 27 = 0$$
Given $$a =2, b=-18, c=-27$$,
1. Use the quadratic root
Use the root solution formula for a quadratic equation, the roots of the equation are given as
$$\begin{aligned} \\x&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-18)\pm\sqrt{(-18)^2-4\cdot 2\cdot (-27)}}{2 \cdot 2}\\ & =\dfrac{18\pm6\sqrt{15}}{4}\\ & =\dfrac{9}{2}\pm\dfrac{3}{2}\sqrt{15}\\ \end{aligned}$$
Since the discriminat is greater than zero, we get two real roots:That is,
$$\begin{cases} x_1 =\dfrac{9}{2}+\dfrac{3}{2}\sqrt{15} \\ x_2=\dfrac{9}{2}-\dfrac{3}{2}\sqrt{15} \end{cases}$$
2. Completing the square method
The idea of completing the square is to transform a quadratic equation to the form of a perfect square equals to a constant, which could be positive or minus. And then solve the equation by taking square root on both sides. For the quadratic equation,
$$2x² - 18x - 27 = 0$$
divide each term by $$2$$ to make the coefficient of the leading term $$1$$.
$$x^2-9x-\dfrac{27}{2}=0$$
Move the constant term $$-\dfrac{27}{2}$$ to the right hand side. Then its sign becomes postive.
$$x^2-9x=\dfrac{27}{2}$$
Add square of the half of $$-9$$, the coefficient of the linear term to both sides.
$$x^2-9x+\Big(\dfrac{9}{2}\Big)^2=\dfrac{27}{2}+\Big(\dfrac{9}{2}\Big)^2$$
Convert the trinomial to the form of perfect square on the left hand side. Sum on the left hand side. Then,
$$\Big(x-\dfrac{9}{2}\Big)^2=\dfrac{135}{4}$$
Taking square roots on both sides of above equation gives
$$\sqrt{\Big(x-\dfrac{9}{2}\Big)^2}=\pm\sqrt{\dfrac{135}{4}}$$
Since the left hand side is square root of a perfect square, we can get rid of radical. Then,
$$x-\dfrac{9}{2}=\pm\dfrac{3}{2}\sqrt{15}$$
Move the constant $$-\dfrac{9}{2}$$ to the right hand side. Then we get,
$$x_1 = 6$$
$$x_2 = 3$$
3. The vertex of the function $$f(x) = 2x² - 18x - 27$$
The vertex of a quadratic function could be determined by completing the square method to transform the quadratic function from general form to vertex form.
For the general form of a quadratic function, we can do the following transformation.
$$\begin{aligned} \\f(x)&=ax^2+bx+c\\ & =a(x^2+\dfrac{b}{a}x)+c\\ & =a\Big[x^2+\dfrac{b}{a}x+\Big(\dfrac{b}{2a}\Big)^2-\Big(\dfrac{b}{2a}\Big)^2\Big]+c\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+c-\dfrac{b^2}{4a}\\ & =a\Big(x+\dfrac{b}{2a}\Big)^2+\dfrac{4ac-b^2}{4a}\\ \end{aligned}$$
Therefore, the corrdinates of the vertex is, $$\Big(\dfrac{-b}{2a},\dfrac{4ac-b^2}{4a}\Big)$$
Here we have, $$a=2$$, $$b=-18$$ and $$c=-27$$. Since $$a >0$$, the curve of the function has the vertex at its lowest point. Substitute them to the vertex formula.
$$\begin{aligned} \\x&=\dfrac{-b}{2a}\\ & =\dfrac{-(-18)}{2\cdot 2}\\ & =\dfrac{9}{2}\\ \end{aligned}$$
$$\begin{aligned} \\y_{min}&=\dfrac{4ac-b^2}{4a}\\ & =\dfrac{4\cdot 2\cdot(-27)-(-18)^2}{4\cdot2^2}\\ & =-\dfrac{135}{2}\\ \end{aligned}$$
So the coordinates for the vertex of the quadrautic function are $$\Big(\dfrac{9}{2},-\dfrac{135}{2}\Big)$$
4. Graph for the function $$f(x) = 2x² - 18x - 27$$
Since the discriminat is greater than zero, the curve of the cubic function $$f(x) = 2x² - 18x - 27$$ has two intersection point with the x-axis