10th Maths
Saturday, 17 March 2018
Saturday, 3 February 2018
Monday, 29 January 2018
Questions of exercise 4.3
Find the roots of the following equations
X X+4 X-7 30
Solution: Let Rehman's present age= x
3 years ago
Rehman's age=x-3
5 years After
Rehman's age= x+5
By the question
1 + 1 = 1
x-3 x+5 3
Cross multiplication
x+5 + x -3 = 1
(x-3)(x+5) 3
2x + 2 = 1
x² + 5x-3x -15 3
2x + 2 = 1
x² +2x -15 3
x² +2x -15 = 6x + 6
x² +2x -15 - 6x - 6 =0
x² - 4x - 21 = 0
Now splitting the middle term
x² -7x +3x -21 = 0
x(x-7) +3(x-7) = 0
(x-7) (x+3) = 0
Either x-7 =0 or x+3=0
=> x=7 or x = -3 (Rejected)
because age can't be negative
Hence Rehman's present age = 7years
Types of polynomials
Polynomials can be divided on the basis of degree and on the basis of number of terms...
On the bases of degree..
On the basis of degree polynomials can be divided into linear,quadratic,cubic,biquadratic, quintic,sextic....e.t.c
LINEAR POLYNOMIAL: A polynomial is said to be linear polynomial if it's degree = 1
e.g (x+1)
Quadratic Polynomial: A polynomial is said to be quadratic polynomial if it's degree =2
e.g x^2 +1....read as (x square plus 1)
CUBIC POLYNOMIAL: A polynomial is said to be cubic polynomial if it's degree =3..
e.g x^3 +1..... read as (x cube plus 1)
BIQUADRATIC POLYNOMIAL: A polynomial is said to be biquadratic polynomial or (Quartic polynomial) if it's degree = 4
e.g x^4 +5...read as x raised to power 4 plus1
On the basis of number of terms
On the basis of number of terms polynomials can be divided into MONOMIALS BINOMIALS TRINOMIALS e.t.c
MONOMIAL: A polynomial is said to be monomial if it contains only one term...MONO means 1
e.g...2x 5y,7z
BINOMIAL:A polynomial is said to be binomial if it contains only two terms...BI means Two.. e.g 2x+3y, 7z+6x etc
TRINOMIAL: A polynomial is said to be trinomial if it contains only three terms...Tri means 3...
e.g 7x+3y+2z
Method of completing squares
Step 1 Divide all terms by a (the coefficient of x square).
Step 2 Move the number term (c/a) to the right side of the equation.
Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. We now have something that looks like (x + p)2 = q, which can be solved rather easily:
Step 4 Take the square root on both sides of the equation. Step 5 Subtract the number that remains on the left side of the equation to find x.
Step 2 Move the number term (c/a) to the right side of the equation.
Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. We now have something that looks like (x + p)2 = q, which can be solved rather easily:
Step 4 Take the square root on both sides of the equation. Step 5 Subtract the number that remains on the left side of the equation to find x.
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