Home Service Math Tutorial

Need of school module assistance? school homework assistance? or week ends face to face tutorial (preferably around Marikina, QC and Cainta Rizal)? DM or PM Mosh L. Adoro for details. Thank you. ❤

Solving the sum and the product of the roots of quadratic equation (ax²+bx+c=0).

To easily compute the sum and the product of the roots of quadratic equation.
Use the formula:
For the sum of roots: -b/a
For the product of roots: c/a

Note: a,b and c are coefficients of quadratic equation and before using the formula make sure na naka general form (ax²+bx+c=0) yung given quadratic equation.

Example 1:
Find the sum and the product of the roots of x²+9x-52=0.

Solution:
Based on the given quadratic equation the value of a,b and c are 1,9 and -52.
By substituting the values to the formula for sum and product of the roots.
The result is:
Sum = -b/a = -(9)/1 = -9
Product = c/a = -52/1 = -52

It means the sum and the product of the roots of quadratic equation x²+9x-52=0 is -9 and -52.

Example 2:
Find the sum and the product of the roots of quadratic equation 2x²-23=10x-11.

Solution:
First, form the given quadratic equation in general form.
2x²-23=10x-11
2x²-23-10x+11=0
2x²-10x-12=0

Now the equation is 2x²-10x-12=0. Based on the equation the value of a,b and c are 2, -10 and -12.

By substituting the values to the formula for sum and product of the roots.
The result is:
Sum = -b/a = -(-10)/2 = 5
Product = c/a = -12/2 = -6

Meaning the sum and the product is 5 and -6.

Try mo nga! 😁
Solve the sum and the product of the roots of the following quadratic equation.
1) x²-5x+6=0
2) 3x²-12x-16=20

~Sir Miks

Solving for the derivative using DELTA METHOD

Note: the symbol like this "∆" read as "Delta".

Problem: What is the derivative of y=2x²+5x+1?

Step1: Substitute x+∆x for x and y+∆y for y:
So y=2x²+5x+1 become y+∆y=2(x+∆x)²+5(x+∆x)+1

Step2: Eliminate y on the left part by subtracting both part by y:
y+∆y=2(x+∆x)²+5(x+∆x)+1
∆y=2(x+∆x)²+5(x+∆x)+1-y

Step3: Substitute the value of y which is 2x²+5x+1 and simplify:
∆y=2(x+∆x)²+5(x+∆x)+1-y
∆y=2(x+∆x)²+5(x+∆x)+1-(2x²+5x+1)
∆y=2(x²+2x∆x+∆x²)+5(x+∆x)+1-2x²-5x-1
∆y=2x²+4x∆x+2∆x²+5x+5∆x+1-2x²-5x-1
∆y=4x∆x+2∆x²+5∆x

Step4: Factor out ∆x from 4x∆x+2∆x²+5∆x then divide both part by ∆x:
∆y=4x∆x+2∆x²+5∆x
∆y=∆x(4x+2∆x+5)

∆y/∆x=∆x(4x+2∆x+5)/∆x
∆y/∆x=4x+2∆x+5

Last Step: Evaluate the limit of ∆y/∆x as ∆x approaches zero ( in other words, substitute zero to ∆x of 4x+2∆x+5):
∆y/∆x=4x+2∆x+5
∆y/∆x=4x+2(0)+5
∆y/∆x=4x+5

It means the derivative of 2x²+5x+1 is 4x+5 😊.

Try mo nga!😁
Solve the derivative of the following using Delta Method.
1) y=x²-100
2) y=2x²+x+18

~Sir Miks