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 *ax*^{2}
+ *bx* +*c* = 0 then,

**Example 3.44**

If the difference between the roots of the equation** ***x*^{2}** **−** **13*x*** **+** ***k*** **=** **0** **is** **17 find *k*.

*Solution*

*x*^{2}* *−* *13*x *+* k *=0* *here,* a *=* *1,*
b *= −13* *,* c *=* k*

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

Also *α* – *β = *17 ….(2)

(1)+(2) we get, 2*α* = 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** ***x*^{2}** **+** **7** ***x*** **+** **10** **=** **0** **find the values of

(i) (*α* - *β*)

(ii) *α*^{2} + *β*^{2}

(iii) *α*^{3} - *β*^{3}

(iv) *α*^{4} + *β*^{4}

(v) *α/β + β/α *

(vi) *α*^{2}*/
β*^{2}* + β*^{2}*/α*^{2}

*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} +* β*^{2}
= (*α* + *β*)^{2} − 2*αβ* = (−7)^{2} − 2 × 10 =
29

(ii) *α*^{3} - *β*^{3 }=
(*α* − *β*)^{3} + 3*αβ*(*α* − *β*) = (3)^{3}
+ 3(10)(3) =117

(ii) *α *^{4}* *+* β*^{4} =
(*α* ^{2} + *β*^{2} )^{2} − 2*α* ^{2}
*β*^{2 }= 29^{2} - 2×(10)^{2} = 641 (since from
(ii)), *α *^{2} +* β*^{2} = 29

**Example 3.46 **

If α , β are the roots of the equation 3*x*^{2 }+
7 *x* − 2 = 0 , find the values of

*Solution*

3*x*^{2} + 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 2*x*^{2} − *x*
−1 = 0 , then form the equation whose roots are

(i) 1/*α* , 1/*β*

(ii) *α* ^{2}*β* , *β* ^{2}*α*

(iii) 2*α* + *β* , 2*β* + *α*

*Solution*

2*x*^{2}** **−

*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 *x*^{2} –(Sum of the roots)*x*
+ (Product of the roots) = 0

*x *^{2}* *−* *(−1)*x *−* *2* *=* *0 gives *x*^{2}*
*+* x *−* *2* *=* *0

(ii) Given roots are *α*^{2} *β* , *β*^{2}
*α*

Sum of the roots *α*^{2}*β*
+ *β*^{2}*α* = *αβ(α+β)* = -1/2(1/2) = -1/4

Product of the roots (*α*^{2}*β*)
× (*β*^{2}*α*) = *α*^{3}*β*^{3} = (*αβ*)^{3}
= (-1/2)^{3} = -1/8

The required equation is *x*^{2} - (Sum of the roots)*x*
+ (Product of the roots)=0

* x*^{2}* *–
(-1/4)* x – *1/8* =* 0 gives 8*x*^{2} + 2*x* – 1 =0

(iii) 2*α* + *β* , 2*β* + *α*

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

Product of the roots = (2*α* + *β* )(2*β* + *α*)
= 4*αβ* + 2*α* ^{2 }+ 2*β*^{2} + *αβ*

The required equation is *x*^{2} - (Sum of the roots)*x*
+ (Product of the roots)=0

*x*^{2}* *–* *3/2* x *+* *0* *=* *0* *gives*
*2*x*^{2}* *−* *3*x *=* *0

Tags : Example Solved Problem | Mathematics Example Solved Problem | Mathematics

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