Urgent help required

[saravananr]

Dear Sir ,

 

I am in IX std . I felt difficulty in solving these problems . So please help me to solve these . I will be very thankful to you .

 

Q 1 . Factorise the following - x^2+x/4-1/8

 

Q.2 . (1+3x)^3 is an example of

a) monomial b)binomial c) trinomial d) none of these

 

 

Thankyou

 

 

[ajb]

Is that [math]- x^{2}+ \frac{x}{4}- \frac{1}{8}[/math] or [math]+ x^{2}+ \frac{x}{4}- \frac{1}{8}[/math] ?

 

The second one with the plus you can factorize (over the reals). My suggestion is pull out the factor of 1/8 and see if you can factorise that. Think about what you have to do and use trial and error if need be.

 

For the second question, tell us the definitions of these terms.

[saravananr]

it is .

 

 

monomial - an algebraic expression with a single term

binomial -an algebraic expression with two terms

trinomial - an algebraic expression with three terms

 

sorry for that error .

 

 

monomial - an algebraic expression with a single term

binomial -an algebraic expression with two terms

trinomial - an algebraic expression with three terms

[ajb]

Expand out the expression and decide what you have.

 

For the factorisation, if you cannot spot it, use The Formula.

[mathematic]

How to think about it. What two rational numbers multiplied together give 1/8? Since the constant has a minus sign, the two numbers must be of opposite sign. Finally they need to add up to 1/4. You should be able to do it quickly.

[Orion1]

Problem:

Factor: [math]x^2 + \frac{x}{4} - \frac{1}{8} = 0[/math]

 

[math]c = \frac{1}{8}[/math]

 

Normalize the equation by multiplying completely through by [math]c^{-1} = 8[/math]

[math]8 \left(x^2 + \frac{x}{4} - \frac{1}{8} \right) = 8(0) = 8x^2 + 2x - 1 = 0[/math]

 

Normalized equation:

[math]8x^2 + 2x - 1 = 0[/math]

 

Manually plot the graph on graph paper:

Plot Graph - Wolframalpha

 

Solve for the normalized coefficients:

[math]a = 8[/math]

[math]b = 2[/math]

[math]c = -1[/math]

 

Use the Quadratic equation:

[math]x = \frac{-b + \sqrt {b^2 - 4ac}}{2a}[/math]

 

[math]x = \frac{-2 + \sqrt {2^2 - 4(8)(-1)}}{2(8)} = \frac{1}{4}[/math]

 

[math]\boxed{x = \frac{1}{4}}[/math]

 

Solve for zero factor:

[math]\left(x - \frac{1}{4} \right) = 0[/math]

 

[math]x = \frac{-b - \sqrt {b^2 - 4ac}}{2a}[/math]

 

[math]x = \frac{-2 - \sqrt {2^2 - 4(8)(-1)}}{2(8)} = - \frac{1}{2}[/math]

 

[math]\boxed{x = - \frac{1}{2}}[/math]

Solve for zero factor:

[math]\left(x + \frac{1}{2} \right) = 0[/math]

 

Combine both zero factors:

[math]\left(x - \frac{1}{4} \right) \left(x + \frac{1}{2} \right) = 0[/math]

 

Factored solution:

[math]\boxed{\left(x - \frac{1}{4} \right) \left(x + \frac{1}{2} \right) = 0}[/math]

---

Problem:

(1+3x)^3 is an example of

a) monomial b) binomial c) trinomial d) none of these

 

[math](ax + 1)^n = 0[/math] - polynomial

[math](ax + 1)^0 = 1[/math] - one non-zero term = monomial

[math](ax + 1)^1 = ax + 1[/math] - two non-zero terms = binomial

[math](ax + 1)^2 = a^2x^2 + 2ax + 1[/math] - three non-zero terms = trinomial

[math](ax + 1)^3 = a^3x^3 + 3a^2x^2 + 3ax + 1[/math] - four non-zero terms = quadnomial

 

The number of non-zero terms is equivalent to the polynomial exponent plus one.

[math]n_t = n + 1[/math]

 

Solution:

[math]\boxed{n = 3}[/math] - quadnomial

 

Answer: d) none of these

Reference:

Quadratic formula - Wikipedia

Plot Graph - Wolframalpha

Polynomials - Wikipedia

Edited September 8, 2012 by Orion1