What method do I study to solve (x squared+3x-4=0)?

Method 1
x² + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = - 4 , x = 1

Method 2
(x² + 3 x + 9/4) - 9/4 - 4 = 0
(x + 3/2)² = 25/4
(x + 3/2) = ± 5/2
x = - 3/2 ± 5/2
x = - 4 , x = 1

Method 3
x = [ - b ± √ (b² - 4 a c ) ] / 2 a
x = [ - 3 ± √ (9 + 16 ) ] / 2
x = [ - 3 ± 5 ] / 2
x = - 4 , x = 1

(3x - 4)(x + 5) = 0
3x - 4 = 0 or x + 5 = 0
x = 3/4 or x = -5

Above two x values are the x values which the curve intercepts x-axis.

To graph it, just make a two columns table: x values and function values.

(3x - 4)(x + 5) = 0
3x - 4 = 0 or x + 5 = 0
x = 3/4 or x = -5


Graph it?
Given one x value, find the function value.
Given another x value, find antoher function value.
Given more more x values, find more values of function

x = 1, y =??
x = 2, y = ? ? ?
x = 3, y = ? /? ?
then, graph it.

Greetings,

Learn a method called "completing the square"

This will allow you to find the vertex of the parabola, by transforming it into the form

f(x) = a(x - h)^2 + k,

if a > 0 this opens upwards if a < 0 this opens downwards
the vertex is at (h,k)
the graph is symmetrical around the line x = h

When f(x) = 0 this graph crosses the x-axis,

The completeing the square method also allows you to calculate these points (if they exist)

For the general quadratic

ax^2 + bx + c = 0
a(x^2 + bx/a) + c = 0
a(x^2 + bx/a + (b/2a)^2) - b^2/4a + c = 0
a(x + b/2a)^2 + (4ac - b^2)/(4a) = 0
a(x + b/2a)^2 = (b^2 - 4ac)/4a
(x + b/2a)^2 = (b^2 - 4ac)/4a^2
x + b/(2a) = +/- sqrt(b^2 - 4ac)/2a
x = -b/(2a) +/- sqrt (b^2 - 4ac)/2a

For f(x) = 3x² + 11x - 20
a = 3, b =11, c = -20

f(x) = a(x + b/2a)^2 - b^2/4a + c from above
f(x) = 3(x + 11/6)^2 - 121/(12) - 20
f(x) = 3(x + 11/6)^2 - 361/12

When f(x) = 0
x = -11/6 +/- sqrt(121 + 12*20)/6

x = -11/6 - 19/6 = -5
or x = -11/6 + 19/6 = 4/3

This parabola opens upwards, has a vertex at (-11/6, -361/12)
and crosses the x-axis at -5 and at 4/3

Regards

There are several methods available, but to study just one method, I would suggest the quadratic formula.

Put the equation into ax² + bx + c = o

The formula is x = (-b +- sqrt(b² - 4ac))/2a

Many times a question will ask, how many solutions there are.

If (b² - 4ac) is positive there are two real solutions.

If (b² - 4ac) is zero, there is one real solution

If (b² - 4ac) is negative there are two imaginary solutions
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You need to study factoring as a first step.
It factors out to (3x - 4)(x + 5) = 0
You can set each factor = to 0 to solve for x.
3x - 4 = 0
3x = 4
x = 3/4 or
x + 5 = 0
x = -5
For graphing set up an x and y table in two columns
Put in several values of x into the original equation. Start with -5, then 0, then 3/4 the answers that you get will be the y values.

solve by factorising or using the quadratic equation.
x^2 + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = 1 or x = -4

to graph this, you know that the curve crossed the y-axis at x = -4 and x = 1, the x^2 coefficient is positive so the curve is negative between x = -4 and x = 1. You can differentiate and equate to zero to find the minimum point:

d/dx = 2x + 3 = 0
2x = -3
x = -3/2
so x = -3/2 at the minimum point, and y = (-3/2)^2 + 3(-3/2) - 4 = -25/4.

You can graph it from that information.
Do the same for the other equation

You can either factor
or use the quadractic formula

Factoring it, you get

(x-1) (x+4)

so x= 1 or x= -4

By using the quadratic formula you will get the same result

For equations in the form of ax^2 + bx + c

-b +- square root of (b^2 - 4ac)
________________________
2a

3x^2 + 11x - 20
becomes
(3x -4) (x+5)
so x= 4/3 or x=-5

There is one way that I did not see. For me this worked also.

X^2 + 3X - 4 = 0
Factor out one X and you get:
X(X + 3 - 4) = 0, X = (-1) Check: 1 + 3 -4 = 0 and
(-1) ^2) + 3 -4 = 0
0 = 0