Answer:
Please see the pictures attached to this answer.
Step-by-step explanation:
A quadratic equation is in the form [tex]ax^2+bx+c=0[/tex] where [tex]x[/tex] represents the unknown and [tex]a\ne{0}[/tex]. The highest degree of a quadratic equation is 2.
1. Factoring method
2. Completing the square method
3. Using the Quadratic Formula
We'll show you the process of solving the solutions of a quadratic equation using the factoring method.
[tex]x^2-7x+10=0[/tex]
Solution:
[tex]\begin{aligned}x^2-7x+10&=0\\x^2-2x-5x+10&=0\\(x^2-2x)+(-5x+10)&=0\\x(x-2)-5(x-2)&=0\\(x-5)(x-2)&=0\\x-5&=0\\x&=5\\&\text{or}\\x-2&=0\\x&=2\end{aligned}[/tex]
Thus, the solution set for the equation [tex]x^2-7x+10=0[/tex] is [tex]\{2, 5\}[/tex].
We'll show you the process of solving the solutions of a quadratic equation using the completing the square method.
[tex]x^2-7x+10=0[/tex]
Solution:
[tex]\begin{aligned}x^2-7x+10&=0\\x^2-7x&=-10\\x^2-7x+\left(\frac{-7}{2}\right)^2&=-10+\left(\frac{-7}{2}\right)^2\\x^2-7x+\frac{49}{4}&=-10+\frac{49}{4}\\x^2-7x+\frac{49}{4}&=\frac{9}{4}\\\left(x-\frac{7}{2}\right)^2&=\frac{9}{4}\\\\\sqrt{\left(x-\frac{7}{2}\right)^2}&=\pm\sqrt{\frac{9}{4}}\\x-\frac{7}{2}&=\pm\frac{3}{2}\\x&=\frac{7}{2}\pm\frac{3}{2}\\x&=\frac{7}{2}+\frac{3}{2}\\x&=\frac{10}{2}\\x&=5\\&\text{or}\\x&=\frac{7}{2}-\frac{3}{2}\\x&=\frac{4}{2}\\x&=2\end{aligned}[/tex]
We'll show you the process of solving the solutions of a quadratic equation using the quadratic formula method.
[tex]x^2-7x+10=0[/tex]
Solution:
[tex]\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(10)}}{2(1)}\\&=\frac{7\pm\sqrt{49-40}}{2}\\&=\frac{7\pm\sqrt{9}}{2}\\&=\frac{7\pm3}{2}\\x&=\frac{7+3}{2}\\x&=5\\&\text{or}\\x&=\frac{7-3}{2}\\x&=2\end{aligned}[/tex]
1. Sum: [tex]r_1+r_2=\frac{-b}{a}[/tex]
2. Product: [tex]r_1\times{r_2}=\frac{c}{a}[/tex]
[tex]x^2-7x+10=0[/tex]
Solution:
The roots of this quadratic equation are [tex]x=2[/tex] and [tex]x=5[/tex]. Note that [tex]a=1[/tex] and [tex]b=-7[/tex]
Sum:
[tex]\begin{aligned}r_1+r_2&\stackrel{?}{=}\frac{-b}{a}\\2+5&\stackrel{?}{=}\frac{-(-7)}{1}\\7&\stackrel{?}{=}\frac{7}{1}\\7&\stackrel{\checkmark}{=}7\end{aligned}[/tex]
The roots of this quadratic equation are [tex]x=2[/tex] and [tex]x=5[/tex]. Note that [tex]a=1[/tex] and [tex]c=10[/tex]
Product:
[tex]\begin{aligned}r_1\times{r_2}&\stackrel{?}{=}\frac{c}{a}\\2\times5&\stackrel{?}{=}\frac{10}{1}\\10&\stackrel{\checkmark}{=}10\end{aligned}[/tex]
Picture 1 indicates that the maximum profit can be achieved at an estimated cost of $225. In this picture, we can see that the parabola opens downward, thus the value of its [tex]a[/tex] is negative.
Picture 2 shows a typical real-life application of the quadratic equation. One person may be able to compute the force to be applied in throwing the ball and the distance from the person to the ring.
Picture 3 shows the bridge having the quadratic equation, how quadratic are illustrated in the pictures is there are several basic ideas used in bridge construction
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