SOLVING EQUATIONS HARDER QUESTIONS

Question 1 :

If

(x/b) + (b/x) = (a/b) + (b/a)

the roots of the equation are

Solution :

(x/b) + (b/x) = (a/b) + (b/a)

(x2+b2)/bx  =  (a2+b2)/ab

ab (x2+b2)  =  bx (a2+b2)

abx2+ab3  =  a2b x + b3x

abx2+ab3-a2bx-b3x  =  0

abx2+(-a2b-b3)x+ab3  =  0

a = ab, b = (-a2b- b3), c = a b3

 x  =  [-b±√(b2-4ac)]/2a  ----(1)

b2-4ac  =  (-a2b- b3)2 - 4(ab)(ab3)

=  (a2b+ b3)- 4(ab)(ab3)

=  a4b2+2a2b4+b6 - 4a2b4

=  a4b2-2a2b4+b6

=  (a2b- b3)2

√(b2-4ac)  =  (a2b- b3)

By applying the value of √(b2-4ac) in (1), we get

x  =  [-(-a2b- b3) ± (a2b- b3)]/2(ab)

x  =  [-(-a2b-b3)+(a2b- b3)]/2ab

(or)

x  =  [-(-a2b- b3) - (a2b- b3)]/2(ab) 

x  =  (a2b+b3+a2b-b3)/2ab

x  =  (a2b+b3+a2b-b3)/2ab

x  =  2a2b/2ab

x  =  a

x  =  [-(-a2b- b3) - (a2b- b3)]/2(ab) 

x  =  (a2b+b3-a2b+b3)/2ab

x  =  2b3/2ab

x  =  b2/a

So, the solution is {a, b2/a}.

Question 2 :

The values of x in the equation

7(x+2p)2+5p2  =  35xp+117 p2

Solution :

7(x+2p)2+5p2  =  35xp+117p2

7(x2+4xp+4p2)+5p2   =  35xp+117p2

7x2+28xp+28p2+5p2  =  35xp+117p2

7x2+28xp-35xp+28p2-117p2+5p2  =  0

7x2-7xp-84p2  =  0

x2-xp-12p2  =  0

x2-4xp+3xp-12p2  =  0

x(x-4p)+3p(x-4p)  =  0

(x+3p)(x-4p)  =  0

By solving, we get

x  =  -3p and x  =  4p

So, the solution is {-3p, 4p}.

Question 3 :

Solving equation

(b-c) x2 + (c-a) x + (a-b)  =  0

Solution :

a = b-c, b  =  c-a, c  =  a-b

x  =  [-b±√(b2-4ac)]/2a  ------(1)

b2-4ac  =  (c-a)2-4(b-c)(a-b)

=  c2+a2-2ac-4(ab-b2-ac+bc)

=  c2+a2-2ac-4ab+4b2+4ac-4bc

=  c2+a2+2ac-4ab+4b2-4bc

=  a2+4b2+c2-4ab-4bc+2ac

=  (a-2b+c)2

√(b2-4ac)  =  a-2b+c

By applying the value of √(b2-4ac) in (1), we get

x  =  [-(c-a) ± (a-2b+c)]/2(b-c)

x  =  [-(c-a)+(a-2b+c)]/2(b-c)

(or)

x  =  [-(c-a)-(a-2b+c)]/2(b-c)

x  =  [-(c-a)+(a-2b+c)]/2(b-c)

x  =  (-c+a+a-2b+c)/2(b-c)

x  =  2(a-b)/2(b-c)

x  =  (a-b)/(b-c)

x  =  [-(c-a)-(a-2b+c)]/2(b-c)

x  =  (-c+a-a+2b-c)/2(b-c)

x  =  2(b-c)/2(b-c)

x  =  1

So, the solution is { (a-b)/(b-c), 1 }.

Question 4 :

The  values of x satisfying the equation

√(2x2+5x-2)-√(2x2+5x-9)  =  1

Solution :

√(2x2+5x-2)-√(2x2+5x-9)  =  1

Taking squares on both sides, we get

(√(2x2+5x-2)-√(2x2+5x-9))2  =  12

(2x2+5x-2)+(2x2+5x-9)-2√(2x2+5x-2)√(2x2+5x-9)  =  1

4x2+10x-11-2√(2x2+5x-2)√(2x2+5x-9)  =  1

4x2+10x-12  =  2√(2x²+5x-2)(2x²+5x-9)

2x2+5x-6  =  √(2x²+5x-2)(2x²+5x-9)

(2x2+5x-6)2  =  (2x²+5x-2)(2x²+5x-9)

(2x2)2+(5x)2+(-6)2+2(2x2)(5x)+2(5x)(-6)+2(-6)(2x2)

           = 4x4+10x3-18x2+10x3+25x2-45x-4x2-10x+18

4x4+25x2+36+20x3-60x-24x2  = 

4x4+20x3-22x2+25x2-45x-10x+18

4x4+25x2+36+20x3-60x-24x2  =  4x4+20x3+3x2-55x+18

4x4-4x4+20x3-20x3+x2-3x2-60x+55x+36-18  =  0

-2x2-5x+18  =  0

2x2+5x-18  =  0

(2x+9) (x-2)  =  0

By solving, we get 

x  =  -9/2 and x  =  2

So, the solution is {-9/2, 2}.

Question 5 :

The  values of x satisfying the equation

Solving equation

6 (√[x/(1-x)]+√[(1-x)/x]] = 15

Solution :

[√x/√(1 - x)] + [√(1 - x)/√x] = 15/6

By taking least common multiple,

(√x√x)+[√(1-x)√(1-x)]/√x(1-x)  =  5/2

x+(1-x)/√(x - x2) = 5/2

2  =  5√(x - x2)

Taking squares on both sides

2 =  (5√(x-x2))2

4  =  25 (x-x2)

4  =  25x-25x2

25x2-25x+4  =  0

25x2-20x-5x+4  =  0

(5x-1) (5x-4)  =  0

x = 1/5 and x = 4/5

Solution is {1/5, 4/5}.

Question 6 :

Solving equation

z2-6z+9  =  4√(z2-6z+6)

following roots are obtained

Solution :

z2-6z+9  =  4√(z2-6z+6)

To remove square root on right side we have to take squares on both sides.

(z2-6z+9)2  =  [4√(z2-6z+6)]²

(z2- 6 z + 9) =  (z2)2+(6z)2+92-2z2(6z) - 2(6z)(9) + 2(9) (z2)

=  z4+36z2+81-12z3-108z+18z2

(z2-6z+9)2  =  [4√(z2-6z+6)]2

[z4+36z2+81-12z3-108z+18z2]  =  16(z2-6z+6)

[z4+36z2+81-12z3-108z+18z2]  =  16z2-96z+96

z4-12z3+38z2-12z-15  =  0

By solving the remaining quadratic equation, we can find other two roots

z2-6z-3  =  0

z  =  [- b ± √(b2-4ac)]/2a

a  =  1, b  =  -6 and c = -3

z  =  [-(-6) ± √(-6)²-4(1)(-3))]/2(1)

z  =  (6±√48)/2

z  =  (6±4√3]/2

z  =  3+2√3 and z  =  3-2√3

Solution is { 1, 5, 3+2√3, 3-2√3 }.

Question 7 :

When

√(2z+1) + √(3z+4)  =  7

the value of z is given by

Solution :

√(2z+1)+√(3z+4)  =  7

By taking squares on both sides, we get

[√(2z+1)+√(3z+4)]2  =  72

[√(2z+1)]2 + [√(3z+4)]2 + 2√(2z+1) (3z+4)  =  49

2z+1+3z+4+2√(6z2+11z+4)  =  49

5z+5+2√(6z2+11z+4)  =  49

2√(6z2+11z+4)  =  49-5z-5

2√(6z2+11z+4)  =  44-5z

By taking squares on both sides, we get

 [2√(6z2+11z+4)w]2  =  (44-5z)2

4 (6z2+11z+4)  =  442-2(44) (5z)+(5z)²

24z2+44z+16  =   1936-440z+25z2

25z2-24z2-440z-44z+1936-16  =  0

z2-484z+1920  =  0

z2-480z-4z+1920  =  0

(z-4) (z-480)  =  0

z  =  4 and z  =  480

solution are x  =  4 and x  =  480.

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