Answer:
x = -3/5 and x = 1.
Explanation:
Certainly! Let's solve each of the quadratic equations step by step.
a) 7x^2 + 19x + 12
To solve this quadratic equation we can use factoring or the quadratic formula. Let's try factoring first.
7x^2 + 19x + 12 = 0
We are looking for two numbers that multiply to 84 (product of the coefficient of x^2 and the constant term) and add up to 19 (coefficient of x). The numbers that satisfy this condition are 4 and 3.
7x^2 + 19x + 12 = (7x + 4)(x + 3) = 0
Setting each factor equal to zero:
7x + 4 = 0 or x + 3 = 0
Solving for x:
7x = -4 or x = -3
x = -4/7 or x = -3
Therefore the solutions are x = -4/7 and x = -3.
b) 6y^2 + 3y - 9
Let's solve this equation using factoring:
6y^2 + 3y - 9 = 0
First we can simplify the equation by dividing every term by 3:
2y^2 + y - 3 = 0
We are looking for two numbers that multiply to -6 (product of the coefficient of y^2 and the constant term) and add up to 1 (coefficient of y). The numbers that satisfy this condition are -2 and 3.
2y^2 - 2y + 3y - 3 = 0
Grouping the terms:
(2y^2 - 2y) + (3y - 3) = 0
Factoring by grouping:
2y(y - 1) + 3(y - 1) = 0
Factoring out the common factor:
(2y + 3)(y - 1) = 0
Setting each factor equal to zero:
2y + 3 = 0 or y - 1 = 0
Solving for y:
2y = -3 or y = 1
y = -3/2 or y = 1
Therefore the solutions are y = -3/2 and y = 1.
c) 5x^2 - 2x - 3
To solve this quadratic equation we can again use factoring or the quadratic formula.
5x^2 - 2x - 3 = 0
We are looking for two numbers that multiply to -15 (product of the coefficient of x^2 and the constant term) and add up to -2 (coefficient of x). The numbers that satisfy this condition are -5 and 3.
5x^2 - 5x + 3x - 3 = 0
Grouping the terms:
(5x^2 - 5x) + (3x - 3) = 0
Factoring by grouping:
5x(x - 1) + 3(x - 1) = 0
Factoring out the common factor:
(5x + 3)(x - 1) = 0
Setting each factor equal to zero:
5x + 3 = 0 or x - 1 = 0
Solving for x:
5x = -3 or x = 1
x = -3/5 or x = 1
Therefore the solutions are x = -3/5 and x = 1.
I hope this helps! Let me know if you have any further questions.