To determine which of the given quadratic equations Ramiya is solving, let's look at the step-by-step solution using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
We need to test each of the given quadratic equations:
1. First Equation: [tex]\( x^2 + 3x + 2 = 0 \)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 2\)[/tex].
The discriminant is:
[tex]\[ b^2 - 4ac = 3^2 - 4(1)(2) = 9 - 8 = 1 \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-3 \pm \sqrt{1}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm 1}{2} \][/tex]
[tex]\[ x_1 = \frac{-3 + 1}{2} = \frac{-2}{2} = -1 \][/tex]
[tex]\[ x_2 = \frac{-3 - 1}{2} = \frac{-4}{2} = -2 \][/tex]
The roots are [tex]\(x_1 = -1\)[/tex] and [tex]\(x_2 = -2\)[/tex].
2. Second Equation: [tex]\( x^2 - 3x + 2 = 0 \)[/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 2\)[/tex].
The discriminant is:
[tex]\[ b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1 \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{3 \pm 1}{2} \][/tex]
[tex]\[ x_1 = \frac{3 + 1}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x_2 = \frac{3 - 1}{2} = \frac{2}{2} = 1 \][/tex]
The roots are [tex]\(x_1 = 2\)[/tex] and [tex]\(x_2 = 1\)[/tex].
3. Third Equation: [tex]\( 2x^2 + 3x + 1 = 0 \)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 1\)[/tex].
The discriminant is:
[tex]\[ b^2 - 4ac = 3^2 - 4(2)(1) = 9 - 8 = 1 \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-3 \pm 1}{4} \][/tex]
[tex]\[ x_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -0.5 \][/tex]
[tex]\[ x_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \][/tex]
The roots are [tex]\(x_1 = -0.5\)[/tex] and [tex]\(x_2 = -1\)[/tex].
4. Fourth Equation: [tex]\( 2x^2 - 3x + 1 = 0 \)[/tex]
Here, [tex]\(a = 2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex].
The discriminant is:
[tex]\[ b^2 - 4ac = (-3)^2 - 4(2)(1) = 9 - 8 = 1 \][/tex]
Therefore, the roots are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-3) \pm \sqrt{1}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{3 \pm 1}{4} \][/tex]
[tex]\[ x_1 = \frac{3 + 1}{4} = \frac{4}{4} = 1 \][/tex]
[tex]\[ x_2 = \frac{3 - 1}{4} = \frac{2}{4} = 0.5 \][/tex]
The roots are [tex]\(x_1 = 1\)[/tex] and [tex]\(x_2 = 0.5\)[/tex].
Given that Ramiya's equation results in the first equation having integer roots of [tex]\(-1\)[/tex] and [tex]\(-2\)[/tex], Ramiya is solving the quadratic equation::
[tex]\[ \boxed{x^2 + 3x + 2 = 0} \][/tex]
