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Example Solved Problem | Mathematics - The Relation between Roots and Coefficients of a Quadratic Equation | 10th Mathematics : UNIT 3 : Algebra

Chapter: 10th Mathematics : UNIT 3 : Algebra

The Relation between Roots and Coefficients of a Quadratic Equation

Mathematics : Algebra: The Relation between Roots and Coefficients of a Quadratic Equation: Example Solved Problem

The Relation between Roots and Coefficients of a Quadratic Equation

Let α and β are the roots of the equation ax2 + bx +c = 0 then,


 

Example 3.44

If the difference between the roots of the equation x2 13x + k = 0 is 17 find k.

Solution

x2 13x + k =0 here, a = 1, b = −13 , c = k

Let  , α, β be the roots of the equation. Then 


Also αβ = 17 ….(2)

(1)+(2) we get, 2α = 30  gives α = 15

Therefore, 15 + β = 13 (from (1)) gives β = −2

But, αβ = c/a = k/1 gives 15 × (−2) = k we get, k = −30

 

Example 3.45

If α and β are the roots of x2 + 7 x + 10 = 0 find the values of

(i) (α - β)

(ii) α2 + β2

(iii) α3 - β3

(iv) α4 + β4

(v) α/β  + β/α

(vi) α2/ β2  + β22

Solution

x 2 + 7 x + 10 = 0 here, a = 1, b = 7 , c =10

If α and β are roots of the equation then,


(ii) α 2 + β2 = (α + β)2 − 2αβ = (−7)2 − 2 × 10 = 29

(ii) α3 - β3 = (αβ)3 + 3αβ(αβ) = (3)3  + 3(10)(3) =117

(ii) α 4 + β4 = (α 2  + β2 )2 − 2α 2 β2 = 292 - 2×(10)2 = 641 (since from (ii)), α 2 + β2 = 29


 

Example 3.46  

If α , β are the roots of the equation 3x2 + 7 x − 2 = 0 , find the values of 


Solution

3x2 + 7 x − 2 = 0 here, a = 3 , b = 7 , c = −2

since, α , β are the roots of the equation


 

Example 3.47

If α , β are the roots of the equation 2x2x −1 = 0 , then form the equation whose roots are

(i) 1/α , 1/β

(ii) α 2β , β 2α

(iii) 2α + β , 2β + α

Solution

2x2 x −1 = 0 here, a = 2 , b = −1 , c = −1 

α + β = −b/a  = −(−1) / 2 = 1/2 ; αβ = c/a = −1/2

(i) Given roots are 1/α , 1/β


The required equation is x2 –(Sum of the roots)x + (Product of the roots) = 0

x 2 (−1)x 2 = 0 gives x2 + x 2 = 0

(ii) Given roots are α2 β , β2 α

Sum of the roots α2β + β2α = αβ(α+β) = -1/2(1/2) = -1/4

Product of the roots (α2β) × (β2α) = α3β3 = (αβ)3 = (-1/2)3 = -1/8

The required equation is x2 - (Sum of the roots)x + (Product of the roots)=0

 x2 – (-1/4) x – 1/8 = 0 gives 8x2 + 2x – 1 =0

(iii) 2α + β , 2β + α

Sum of the roots  2α + β + 2β + α = 3(α + β) = 3(1/2) = 3/2

Product of the roots = (2α + β )(2β + α) = 4αβ + 2α 2 + 2β2 + αβ


The required equation is x2 - (Sum of the roots)x + (Product of the roots)=0

x2 3/2 x + 0 = 0  gives 2x2 3x = 0


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