1) Find the nature of roots for the following quadratic equations. The nature of roots is determined by the discriminant $\Delta = b^2 - 4ac$. • If $\Delta > 0$, the roots are real and distinct. • If $\Delta = 0$, the roots are real and equal. • If $\Delta < 0$, the roots are complex (non-real) and distinct. a) $2x^2 - 5x + 2 = 0$ Step 1: Identify the coefficients. For $2x^2 - 5x + 2 = 0$, we have $a = 2$, $b = -5$, $c = 2$. Step 2: Calculate the discriminant. $$ \Delta = b^2 - 4ac = (-5)^2 - 4(2)(2) $$ $$ \Delta = 25 - 16 $$ $$ \Delta = 9 $$ Step 3: Determine the nature of the roots. Since $\Delta = 9 > 0$, the roots are real and distinct. The roots are $\boxed{\text{real and distinct}}$. b) $3x^2 - 7x + 4 = 0$ Step 1: Identify the coefficients. For $3x^2 - 7x + 4 = 0$, we have $a = 3$, $b = -7$, $c = 4$. Step 2: Calculate the discriminant. $$ \Delta = b^2 - 4ac = (-7)^2 - 4(3)(4) $$ $$ \Delta = 49 - 48 $$ $$ \Delta = 1 $$ Step 3: Determine the nature of the roots. Since $\Delta = 1 > 0$, the roots are real and distinct. The roots are $\boxed{\text{real and distinct}}$. 2) If $\alpha$ and $\beta$ are the roots of the equation $3x^2 - 5x + 4 = 0$, find the value of $\alpha/\beta + \beta/\alpha$. Step 1: Identify the coefficients and find the sum and product of the roots. For $3x^2 - 5x + 4 = 0$, we have $a = 3$, $b = -5$, $c = 4$. Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{(-5)}{3} = \frac{5}{3}$. Product of roots: $\alpha\beta = \frac{c}{a} = \frac{4}{3}$. Step 2: Simplify the expression $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$. $$ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} $$ We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substitute this into the expression: $$ \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} $$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha\beta$. $$ \frac{\left(\frac{5}{3}\right)^2 - 2\left(\frac{4}{3}\right)}{\frac{4}{3}} $$ $$ = \frac{\frac{25}{9} - \frac{8}{3}}{\frac{4}{3}} $$ $$ = \frac{\frac{25}{9} - \frac{24}{9}}{\frac{4}{3}} $$ $$ = \frac{\frac{1}{9}}{\frac{4}{3}} $$ $$ = \frac{1}{9} \times \frac{3}{4} $$ $$ = \frac{3}{36} $$ $$ = \frac{1}{12} $$ The value is $\boxed{\frac{1}{12}}$. 3) If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 4x + 7 = 0$, find $\alpha^2 + \beta^2$. Step 1: Identify the coefficients and find the sum and product of the roots. For $2x^2 - 4x + 7 = 0$, we have $a = 2$, $b = -4$, $c = 7$. Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{(-4)}{2} = \frac{4}{2} = 2$. Product of roots: $\alpha\beta = \frac{c}{a} = \frac{7}{2}$. Step 2: Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substitute the values of $\alpha + \beta$ and $\alpha\beta$: $$ \alpha^2 + \beta^2 = (2)^2 - 2\left(\frac{7}{2}\right) $$ $$ = 4 - 7 $$ $$ = -3 $$ The value is $\boxed{-3}$. 4) Construct and simplify equations whose roots are $-3, 1$. Step 1: Find the sum and product of the roots. Given roots are $r_1 = -3$ and $r_2 = 1$. Sum of roots: $S = r_1 + r_2 = -3 + 1 = -2$. Product of roots: $P = r_1r_2 = (-3)(1) = -3$. Step 2: Form the quadratic equation using the formula $x^2 - Sx + P = 0$. $$ x^2 - (-2)x + (-3) = 0 $$ $$ x^2 + 2x - 3 = 0 $$ The equation is $\boxed{x^2 + 2x - 3 = 0}$.