ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 aLet's start by rewriting the given equation so there are no parentheses and all of the terms are on the left-hand side of the equation.
Now we can identify the values of a, b, and c.
3x^2-4x-5=0 ⇕ 3x^2+( - 4)x+( - 5)=0
We see that a= 3, b= - 4, and c= - 5. Let's substitute these values into the Quadratic Formula.
We found that the solutions of the given equation are x_1= 2+sqrt(19)3 and x_2= 2-sqrt(19)3.
Recall that any square root where the radicand is not a perfect square, such as 4, 9, or 16, will have infinite non-repeating decimals. Since 19 is not a perfect square, and we cannot rewrite the fractions, the solutions are irrational numbers.
b We want to solve the given quadratic equation. To do this, we will take the square root of both sides of the given equation. Since this method gives two solutions — a negative and a positive — remember to consider them both by adding ± to the solution.
The solutions for this equation are x= 4 ± 16. Let's separate them into the positive and negative cases.
x=4 ± 1/6
x_1=4 + 1/6
x_2=4 - 1/6
x_1=5/6
x_2=3/6
x_1=5/6
x_2=1/2
We found that the solutions of the given equation are x_1= 56 and x_2= 12. Recall that a rational number is a number that can be written as a fraction ab, where a and b are integers. Since this is the case, we can see that our solutions are rational.
