Solve the following quadratic equations and find the discriminants: 1) 2x^2 + 7x + 3 = 0, 2) x^2 - 8x + 16 = 0, 3) 2x^2 - 4x + 5 = 0.

Steps to Solve

1. Calculate the Discriminant (D) for the First Equation The first equation is $2x^2 + 7x + 3 = 0$. The discriminant is calculated as follows: $$ D = b^2 - 4ac $$ Here, $a = 2$, $b = 7$, and $c = 3$. Thus, $$ D = 7^2 - 4 \cdot 2 \cdot 3 = 49 - 24 = 25 $$

2. Determine the Roots for the First Equation Now use the quadratic formula: $$ x = \frac{-b \pm \sqrt{D}}{2a} $$ Substituting in the values: $$ x_{1,2} = \frac{-7 \pm \sqrt{25}}{2 \cdot 2} = \frac{-7 \pm 5}{4} $$ This leads to two potential solutions: $$ x_1 = \frac{-2}{4} = -\frac{1}{2} $$ $$ x_2 = \frac{-12}{4} = -3 $$

3. Calculate the Discriminant (D) for the Second Equation The second equation is $x^2 - 8x + 16 = 0$. Here, $a = 1$, $b = -8$, and $c = 16$. Calculate the discriminant: $$ D = (-8)^2 - 4 \cdot 1 \cdot 16 = 64 - 64 = 0 $$ Since $D = 0$, this indicates one repeated root.

4. Determine the Root for the Second Equation Using the quadratic formula, we find: $$ x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1} = \frac{8}{2} = 4 $$ Thus, the root is $x = 4$.

5. Calculate the Discriminant (D) for the Third Equation The third equation is $2x^2 - 4x + 5 = 0$. Here, $a = 2$, $b = -4$, and $c = 5$. The discriminant is: $$ D = (-4)^2 - 4 \cdot 2 \cdot 5 = 16 - 40 = -24 $$ A negative discriminant indicates no real roots.