Answer:
The correct option is: A. [tex]0=x^2+3x+2[/tex]
Step-by-step explanation:
Quadratic formula: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
After substituting the values of a, b and c into the formula, Ramiya's equation is: [tex]x=\frac{-3\pm\sqrt{3^2-4(1)(2)}}{2(1)}[/tex]
Comparing the above equation with the quadratic formula, we will get........
[tex]2a= 2(1)\\ a=1\\ \\ -b=-3\\ b=3\\ \\ 4ac=4(1)(2)\\ 4(1)c=4(1)(2)\\ c=2[/tex]
The standard form of any quadratic equation is: [tex]ax^2+bx+c=0[/tex]
So, Ramiya’s quadratic equation will be: [tex]1x^2+3x+2=0[/tex] or, [tex]0=x^2+3x+2[/tex]
The Ramya's quadratic equation is 0 = x^2 + 3x + 2 option (A) is correct.
Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.
The quadratic formula is:
[tex]\rm x=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
Ramiya equation is:
[tex]\rm x=\frac{-3\pm \sqrt{3^2-4(1)(2)} }{2(1)}[/tex]
Comparing the Ramiya equation to the quadratic formula, we get:
-b = -3 ⇒ b = 3
2a = 2(1) ⇒ a = 1
4ac = 4(1)(2) ⇒ c = 2
Put these values in the standard equation, we get:
[tex]\rm x^2+3x+2=0[/tex] or
[tex]0=\rm x^2+3x+2[/tex]
Thus, the Ramya's quadratic equation is 0 = x^2 + 3x + 2 option (A) is correct.
Learn more about quadratic equations here:
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