Question

use the discriminant to tell how many and the nature of each solution, then solve for

1. $3x^2 - 27x + 60 = 0$
2. $2x^2 + 5x = 3$

Explanation:

Step1: Identify a,b,c for Q11

For $3x^2 -27x +60=0$, $a=3$, $b=-27$, $c=60$

Step2: Calculate discriminant for Q11

Discriminant formula: $\Delta = b^2 -4ac$
$\Delta = (-27)^2 -4(3)(60) = 729 - 720 = 9$

Step3: Analyze Q11 discriminant

$\Delta=9>0$, so 2 real rational solutions.

Step4: Solve Q11 via quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{\Delta}}{2a}$
$x=\frac{27\pm\sqrt{9}}{2(3)}=\frac{27\pm3}{6}$

Step5: Compute Q11 solutions

$x_1=\frac{27+3}{6}=\frac{30}{6}=5$, $x_2=\frac{27-3}{6}=\frac{24}{6}=4$

Step6: Rewrite Q12 to standard form

$2x^2 +5x -3=0$, so $a=2$, $b=5$, $c=-3$

Step7: Calculate discriminant for Q12

$\Delta = 5^2 -4(2)(-3)=25 +24=49$

Step8: Analyze Q12 discriminant

$\Delta=49>0$, so 2 real rational solutions.

Step9: Solve Q12 via quadratic formula

$x=\frac{-5\pm\sqrt{49}}{2(2)}=\frac{-5\pm7}{4}$

Step10: Compute Q12 solutions

$x_1=\frac{-5+7}{4}=\frac{2}{4}=\frac{1}{2}$, $x_2=\frac{-5-7}{4}=\frac{-12}{4}=-3$

Answer:

For $3x^2 -27x +60=0$:

• Number/nature of solutions: 2 distinct real rational solutions
• Solutions: $x=4$ and $x=5$

For $2x^2 +5x=3$ (rewritten as $2x^2 +5x -3=0$):

• Number/nature of solutions: 2 distinct real rational solutions
• Solutions: $x=-3$ and $x=\frac{1}{2}$