MANIPULATING EXPRESSIONS INVOLVING ALPHA NAD BETA

Question 1 :

If α and β are the roots of the equation

3x²-5x+2  =  0

then find the values of

(i) (α/β) + (β/α)

(ii) α - β

(iii) (α²/β) + (β²/α)

Solution :

By comparing the given quadratic equation with the general form of quadratic equation, we get

a  =  3, b  =  -5 and c = 2

(i)  (α/β)+(β/α)

By combining the above fractions, we get

(α/β)+(β/α)  =  (α²+β²)/αβ  -----(1)

α²+β²  =  (α+β)²-2αβ

=  (5/3)²-2(2/3)

=  (25/9)-(4/3)

=  (25-12)/9

α²+β²  =  13/9

By applying the values in (1), we get

 (α/β) + (β/α)  =  (α²+β²)/αβ

=  (13/9)/(2/3)

 (α/β) + (β/α)  =  13/6

(ii) α - β

α-β  =  √(α+β)² - 4αβ

=  √(5/3)²-4(2/3)

=  √(25/9)-(8/3)

=  √1/9

α-β  =  ± 1/3

(iii)  (α²/β) + (β²/α)

By combining the given fractions, we get

(α²/β) + (β²/α)  =  (α³+β³)/αβ  ----(1)

α³-β³  =  (α-β)³ + 3αβ(α-β)

=  (5/3)³-3(2/3)(5/3)

=  (125/27)-(10/9)

α³-β³  =  95/27

By applying the values in (1), we get

(α²/β) + (β²/α)  =  (95/27)/(2/3)

(α²/β) + (β²/α)  =  95/18

Question 2 :

If α and β are the roots of

3x²-6x+4  =  0

find the value of α²+β²

Solution :

 a = 3  b = - 6 and c = 4

α²+β²  =  (α+β)² - 2αβ  ----(1)

By applying the values in (1), we get

=  2²-2(4/3)

=  4-(8/3)

=  4/3

Question 3 :

If roots of the equation x² + x + r = 0 are α and β and α³+β³ = -6. Find the value of r.

Solution :

x² + x + r = 0

a = 1, b = 1 and c = 1

α³+β³ = -6

α³+β³ = (α + β)³ - 3αβ (α + β)

(α + β) = -b/a and α β = c/a

(α + β) = -1/1 and α β = r/1

α³+β³ = (-1)³ - 3(r) (1)

-6 = -1 - 3r

-6 + 1 = 3r

3r = -5

r = -5/3

So, the value of r is -5/3.

Question 4 :

If difference between the roots of the equation x² - kx + 8 = 0 is 4, then find the value of k.

Solution :

For any quadratic equation, which is in the form ax² + bx + c = 0 the roots are α and β

Sum of the roots α + β = -b/a

Product of roots α β = c/a

Difference of the roots

α - β = √[(α + β)² - 4 α β]

 x² - kx + 8 = 0

a = 1, b = -k and c = 8

α + β = k/1 ==> k

α β = 8/1 ==> 8

α - β = k² - 4 (8)

Applying the value of α - β, we get

4 = √(k² - 32)

(k² - 32) = 16

k² = 32 + 16

k² = 48

k = √48

k = √2 · 2 · 2 · 2 · 3

k = (2 · 2)√3

k = ± 4√3

k = 4√3 and -4√3

So, the values of k are -4√3 and 4√3.

Question 5 :

If one of the roots of the equation

x² + px + a = 0 is √3+2

then find the value of p and a is.

Solution :

Every quadratic equation will have two roots α and β. If rational number will be one of the given roots, its conjugate will be the other root.

For example,

if  a + √b is one of the roots then its conjugate  a - √b will be other root.

Since for the given equation 2 + √3 be one root, then its conjugate 2 - √3 will be the other root.

α = 2 + √3 and β = 2 - √3

Sum of roots :

α + β = 2 + √3 + 2 - √3 ==> 4

Product of roots :

α  β = (2 + √3)  (2 - √3)

= 2² - √3²

= 4 - 3

α  β = 1

From the equation,

x² + px + a = 0

α + β = -p/1 ==> -p

4 = -p

p = -4

So, the value of p is -4

From the equation,

x² + px + a = 0

α β = a/1 ==> a

1 = a

So, the value of a is 1.

Question 6 :

If one root of the equation 

px² + qx + r = 0 

is r then other roots of the equation will be

a)  1/q      b)  1/r      c)  1/p      d) 1/(p + q)

Solution :

px² + qx + r = 0

One root of the equation is α = r, then β = ?

From the given equation, we get

Sum of roots = α + β = -q/p ---(1)

Product of roots = α β = r/p ----(2)

Applying the value α = r in (1) and (2), we get

r + β = -q/p

r β = r/p

β = 1/p

So, the other roots is 1/p. Option c is correct.

Question 7 :

One root of the equation 

x² - 2(5 + m) x + 3(7 + m) = 0

is reciprocal of the other. Find the value of m.

a)  -7      b)  7      c)  1/7      d) -1/7

Solution :

x² - 2(5 + m) x + 3(7 + m) = 0

x² - (10 - 2m) x + (21 + 3m) = 0

Sum of roots = α + β = (10 - 2m)/1 ---(1)

Product of roots = α β = (21 + 3m)/1 ----(2)

α and 1/α are the roots.

 α (1/α) = (21 + 3m)

1 = 21 + 3m

1 - 21 = 3m

m = -20/3

m = -6.6

Approximately the value of m is -7.

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