Sure, let's determine the roots of the given equation [tex]\(2x(x-3)=3x+5\)[/tex] by completing the square. Here is the step-by-step solution:
1. Rewrite the equation in standard quadratic form:
Start with the given equation:
[tex]\[
2x(x-3) = 3x + 5
\][/tex]
Expand and simplify the left-hand side:
[tex]\[
2x^2 - 6x = 3x + 5
\][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[
2x^2 - 6x - 3x - 5 = 0
\][/tex]
Combine like terms:
[tex]\[
2x^2 - 9x - 5 = 0
\][/tex]
2. Identify the coefficients:
For the equation [tex]\(2x^2 - 9x - 5 = 0\)[/tex], the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = -9\)[/tex]
- [tex]\(c = -5\)[/tex]
3. Complete the square:
To complete the square, we need to transform the quadratic equation into a perfect square form. Here’s the detailed process:
a. Move the constant term to the other side:
[tex]\[
2x^2 - 9x = 5
\][/tex]
b. Divide the entire equation by the leading coefficient [tex]\(a = 2\)[/tex]:
[tex]\[
x^2 - \frac{9}{2}x = \frac{5}{2}
\][/tex]
c. To complete the square, take half of the coefficient of [tex]\(x\)[/tex], square it, and add it to both sides:
The coefficient of [tex]\(x\)[/tex] is [tex]\(-\frac{9}{2}\)[/tex]. Half of this is [tex]\(-\frac{9}{4}\)[/tex]. Squaring it, we get:
[tex]\[
\left(-\frac{9}{4}\right)^2 = \frac{81}{16}
\][/tex]
Add [tex]\(\frac{81}{16}\)[/tex] to both sides:
[tex]\[
x^2 - \frac{9}{2}x + \frac{81}{16} = \frac{5}{2} + \frac{81}{16}
\][/tex]
d. Simplify the right side:
Convert [tex]\(\frac{5}{2}\)[/tex] to a fraction with a denominator of 16:
[tex]\[
\frac{5}{2} = \frac{40}{16}
\][/tex]
Adding the fractions:
[tex]\[
\frac{40}{16} + \frac{81}{16} = \frac{121}{16}
\][/tex]
Now we have:
[tex]\[
x^2 - \frac{9}{2}x + \frac{81}{16} = \frac{121}{16}
\][/tex]
e. Left side is a perfect square:
[tex]\[
\left( x - \frac{9}{4} \right)^2 = \frac{121}{16}
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
Take the square root of both sides:
[tex]\[
x - \frac{9}{4} = \pm \frac{\sqrt{121}}{4}
\][/tex]
[tex]\[
x - \frac{9}{4} = \pm \frac{11}{4}
\][/tex]
5. Consider both cases for solutions:
a. [tex]\( x - \frac{9}{4} = \frac{11}{4} \)[/tex]:
[tex]\[
x = \frac{9}{4} + \frac{11}{4} = \frac{20}{4} = 5
\][/tex]
b. [tex]\( x - \frac{9}{4} = -\frac{11}{4} \)[/tex]:
[tex]\[
x = \frac{9}{4} - \frac{11}{4} = \frac{-2}{4} = -\frac{1}{2}
\][/tex]
So, the roots of the given equation [tex]\(2x(x-3) = 3x + 5\)[/tex] are:
[tex]\[
x = 5 \quad \text{or} \quad x = -\frac{1}{2}
\][/tex]
