Question

When the sum of 5 and three times a positive number is subtracted from the square of the number, 0 results. Find the number.

Answer 1

To solve for the positive number when the sum of 5 and three times the number is subtracted from the square of the number to yield zero, set up and solve the quadratic equation x^2 - 3x - 5 = 0 using the quadratic formula.

Step-by-step explanation:

The student has asked for help in finding a positive number when the sum of 5 and three times the number is subtracted from the square of the number, and the result is 0. This involves setting up a quadratic equation and solving for the number.

Let's denote this positive number as 'x'. According to the problem statement, the equation can be written as:

x2 - (5 + 3x) = 0.

We can then expand and simplify this equation:

x2 - 5 - 3x = 0

And, rearrange it into standard quadratic form:

x2 - 3x - 5 = 0

Next, we can either factorize this equation, if possible, or use the quadratic formula to find the value of 'x' that satisfies the equation. Since this equation does not factor neatly, we use the quadratic formula:

x = ∛2 - 4ac) / 2a

In this case, a = 1, b = -3, and c = -5. Plugging these into the quadratic formula, we find the two possible solutions for x. Only the positive solution is relevant since the problem states that the number is positive.

Answer 2

`x-the\ number\\\\x^2-(5+3x)=0\\\\x^2-5-3x=0\\\\x^2-3x-5=0`


`Use\ the\ quadratic\ formula:\\a^2x+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ x_1=(-b-\sqrt\Delta)/(2a)\ and\ x_2=(-b+\sqrt\Delta)/(2a)\\\\if\ \Delta=0\ then\ x_0=(-b)/(2a)\\\\if\ \Delta < 0\ then\ no\ solution`


`x^2-3x-5=0\\\\a=1;\ b=-3;\ c=-5\\\\\Delta=(-3)^2-4\cdot1\cdot(-5)+3+20=29 \ \textgreater \ 0\\\\\sqrt\Delta=√(29)\\\\x_1=(3-√(29))/(2\cdot1)=(3-√(29))/(2) \ \textless \ 0\\\\x_2=(3+√(29))/(2\cdot1)=(3+√(29))/(2) \ \textgreater \ 0`


`Answer:\ \boxed{x=(3+√(29))/(2)}`