Math help? x^2=3x+10?

5-2

Question Number 1 :
For this equation x^2 = 3*x + 10 , answer the following questions :
A. Find the roots using Quadratic Formula !

Answer Number 1 :
First, we have to turn equation : x^2 = 3*x + 10 , into a*x^2+b*x+c=0 form.
x^2 = 3*x + 10 , move everything in the right hand side, to the left hand side of the equation
<=> x^2 - ( 3*x + 10 ) = 0 , which is the same with
<=> x^2 + ( - 3*x - 10 ) =0 , now open the bracket and we get
<=> x^2 - 3*x - 10 = 0

The equation x^2 - 3*x - 10 = 0 is already in a*x^2+b*x+c=0 form.
As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = -3, c = -10.

1A. Find the roots using Quadratic Formula !
Use the formula,
x1 = (-b+sqrt(b^2-4*a*c))/(2*a) and x2 = (-b-sqrt(b^2-4*a*c))/(2*a)
We had know that a = 1, b = -3 and c = -10,
then the value a,b and c in the abc formula, can be subtituted.
So we get x1 = (-(-3) + sqrt( (-3)^2 - 4 * (1)*(-10)))/(2*1) and x2 = (-(-3) - sqrt( (-3)^2 - 4 * (1)*(-10)))/(2*1)
Which is the same with x1 = ( 3 + sqrt( 9+40))/(2) and x2 = ( 3 - sqrt( 9+40))/(2)
Which make x1 = ( 3 + sqrt( 49))/(2) and x2 = ( 3 - sqrt( 49))/(2)
So we get x1 = ( 3 + 7 )/(2) and x2 = ( 3 - 7 )/(2)
The answers are x1 = 5 and x2 = -2

x^2 - 3 x - 10 = 0

( x - 5 ) ( x + 2 ) = 0

x = 5 , x = - 2

x^2 + 3x > 10
x^2 + 3x - 10 > 0

Using factorisation you get two numbers that when multiplied give -10 and when added give 3. Two such numbers would be 5 & -2.

(x + 5)*(x - 2) > 0
x > -5 & x > 2.
To satisfy these inequalities x > 2.

move stuff around, to get:

x^2 - 3x - 10 = 0

Which is one standard form for a "quadratic" equation.

There is a recipe, called the "quadratic formula".

Standard form:

a x^2 + bx + c = 0
in your problem, a=1, b= -3 and c = -10 (be careful with signs)

recipe:

x = [ -b +/- sqrt( b^2 - 4ac ) ] / 2a

x = [ +3 +/- sqrt( 9 -4(1)(-10) ) ] / 2(1)
x = [ 3 +/- sqrt (9 + 40) ] / 2
x = [3 +/- sqrt(49)] / 2

+/- means "plus or minus" as there are usually two answers.
Begin with +

x = [ 3 + 7] / 2 = 10/2 = 5

x = [3 - 7] / 2 = -4/2 = -2

These values are called the "roots" of the quadratic, also the "zeros" of the equation (both names mean the same thing)

x²=3x+10
x²-3x-10=0
The discriminant is : (-3)²-4.1.(-10)=49>0
There are two real roots.
x=(3+7)/2 and x=(3-7)/2
x=5 and x=-2

x^2=3x+10
x^2-3x-10=0
(x-5)(x+2)=0
x-5=0 OR x+2=0
x=5 x=-2

x^2 = 3x + 10
x^2 - 3x - 10 = 0
x^2 + 2x - 5x - 10 = 0
(x^2 + 2x) - (5x + 10) = 0
x(x + 2) - 5(x + 2) = 0
(x + 2)(x - 5) = 0

x + 2 = 0
x = -2

x - 5 = 0
x = 5

∴ x = -2, 5

x²= 3x+10

x²-3x-10=0

(x-5)(x+2)=0

x= 5 OR -2

x^2-3x-10=0
factorise - two numbers that times to make -10 and add to make -3
therefore, -5 and 2

(x-5)(x+2)=0
so x-5=0, x+2=0

x=5, -2.