Solve Quadratic Equation Q1: x^2+4x-5=0

[Danatron]

Question 1

(a) Solve the quadratic equation

x^2 + 4x -5 = 0

(b) Factorise its left hand side.

(c) Find interval(s) of x where the left hand side is positive


Q1
(a)
(x+-1) (x+5)
x= 1 x=-5
is that solving the equation?

(b)
didnt i already factorise the left hand side of the equation when i solved for in (a)?
(x+-1)?

(c)
not sure on this one

 

[Jilang]

Hi there, you have solved the equation in a). Yes you did already factorise the equation, but perhaps the question assumed you might have solved it a different way (by the quadratic equation). For part c you need to use the fact the two negative numbers multiplied together will give you a positive number etc.

 

[Ray Vickson]

Question 1

(a) Solve the quadratic equation

x^2 + 4x -5 = 0

(b) Factorise its left hand side.

(c) Find interval(s) of x where the left hand side is positive


Q1
(a)
(x+-1) (x+5)
x= 1 x=-5
is that solving the equation?

(b)
didnt i already factorise the left hand side of the equation when i solved for in (a)?
(x+-1)?

(c)
not sure on this one

Your notation x+-1 is meaningless, and should never be used. If x=1 is a root then one of the factors is x-1; if x = -1 is a root, one of the factors is x+1.

 

[Mentallic]

You should try graphing the quadratic

y=x^2+4x-5

You should know that the quadratic is convex (in the shape of a u) instead of concave (in the shape of an n) because the coefficient of x² (multiple of x²) is 1 and hence positive (if it were negative the quadratic would be concave).

You've also found that it crosses the x-axis at x=1 and x=-5 because this is where the quadratic =0 hence y=0 for those x values.

Putting these two ideas together, it's quite obvious what the domain is (values of x) for where the quadratic is positive.

 

[NascentOxygen]

Q1
(a)
(x+-1) (x+5)
x= 1 x=-5
is that solving the equation?

Yes, you have found the values of the unknowns which will make the right side equal to the left side.

(c)
not sure on this one

You are trying to determine your answer without sketching the graph? ✻shame✻

Always sketch the graph, so you can see what you are dealing with.

 

[mafagafo]

For c), besides sketching the graph you can analyse the sign of the terms individually.

(x-1)(x+5)

When is (x-1) positive? When is it negative?
Same to (x+5).

What must be the signs of (x-1) and (x+5) so that (x-1)(x+5) > 0?

 

[Danatron]

quadratic

ok so i have graphed the quadratic and circled in red all the positive values of x, how would i write this?

[PLAIN]http://i.imgur.com/WwvsteW.jpg?1[/PLAIN]

 

[SammyS]

ok so i have graphed the quadratic and circled in red all the positive values of x, how would i write this?

[PLAIN]http://i.imgur.com/WwvsteW.jpg?1[/PLAIN]

What you're asked for is:

All of the x values for which the y values are positive .

 

[Danatron]

so that would be everything before -5 and everything after 1 right?

 

[Matterwave]

so that would be everything before -5 and everything after 1 right?

Yes. So just turn that into a mathematical expression. What's "everything before -5" and "everything after 1" in mathematical terms?

 

[Danatron]

x < -5 & x > 1

 

[thelema418]

Question 1

(a) Solve the quadratic equation

x^2 + 4x -5 = 0

(b) Factorise its left hand side.

(c) Find interval(s) of x where the left hand side is positive


Q1
(a)
(x+-1) (x+5)
x= 1 x=-5
is that solving the equation?

(b)
didnt i already factorise the left hand side of the equation when i solved for in (a)?
(x+-1)?

(c)
not sure on this one

hmm. In part (a), "solve" does seem confusing in this problem. There are many other ways of solving. Sometimes textbooks want you to graph first. You could also use a rational root test, or something like that. Some books give problems with $ax^2 + bx + c = N$. The concept of "solving" is to graph this as two functions and find their intersection points. Of course, it seems like a lot more work than necessary.

There is a concern I have with what you wrote in part (a). The expression $(x + -1)(x+5)$ does not mean $x = 1$ or $x = -5$. You have to state $(x + -1)(x+5) = 0$ to find those solutions for $x$. I think you KNOW that you are doing this, you just didn't write it. If you did that on a test, a teacher might take points off.

This is also a big idea that connects with part (c). In part (a) you are using the zero property: Zero times a number is zero. In part (c) you are using the following concepts about multiplication of real numbers: "a positive times a positive is a positive"; "a negative times a negative is a positive"; etc. Again these positive & negative multiplication rules are for real numbers, not the imaginary numbers.