Solution

Hints for difficult problems

1. If P & q are the roots of the equation $2a^{2}-4a+1=0$ find the value of

$p^{3}+q^{3}$
Pre requisites:

1. Standard form of quadratic equation
2. Formula to find the sum & product of quadratic equation
3. Knowledge of using appropriate identity

Interpretation of the Problem:

1. Compare the equation with standard form and identify the values of a,b,c
2. To find the sum formformof the roots of the quadratic equation using the formula
3. To find the product of the roots of the equation
4. Using the identity & rewriting $p^{3}+q^{3}$ as $(p+q)^{3}-3pq(p+q)$
5. Substitute the values of m+n & mn in $(p+q)^{3}-3pq(p+q)$
6. Simplification

Concepts:

1. Formula to find the sum and product of the roots of the quadratic equation
2. Identity $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)$

Algorithm:
Consider the equation $2a^{2}-4a+1=0$
Here a=2,b=-4 & c=1
If p & q are the roots of the quadratic equation then
$p+q={\frac {-b}{a}}={{\frac {-(-4)}{2}}=2}$
$pq={\frac {c}{a}}={\frac {1}{2}}$
Therefore,
$p^{3}+q^{3}=(p+q)^{3}-3pq(p+q)$
=$(2)^{3}-3[{\frac {1}{2}}](2)$
=8-3
=5
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2.The altitude of a triangle is 6cm greter than its base. If its area is 108cmsq .Find its base.
solution
Statement: Solving problem based on quadratic equations.
solution

• Interpretation of the problem:
  * Converting data in to eqn.
  *Knowledge about area of a triangle.
  *knowledge of the formula of area of triangle.
  *Methods of finding the roots of the eqn.
  *Methods of finding the roots of the
• Different approches to solve the problem:
  *Factorisation
• Using formula
• using graph
• Concept used:Forming the eqn. 216=x(x+6)

216=x2+6x
x2 +6x -216=0
Substitution: x 2 +18x-12x -216=0
Simplification: x(x+18)-12(x+18)=0
(x+18)( x-12)=0
(x+18)=0 (x-12)=0
x=-18, x=12
.

1. Base=12cm,
   Altitude=x+6

=12+6=18cm.
Prior Knowledge -

• Methods of solving the Eqn
  
• Factorisation
  
• Using Formula
  
• Using Graph
  

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