PROBLEMS ON QUADRATIC EQUATIONS

Problem 1 : 

Find the value of

Solution :

Then,

The quadratic equation above can not be solved by factoring. So, we can use quadratic formula and solve.

Comparing ax2 + bx + c = 0 and x2 - 4x - 1 = 0,

a = 1, b = -4 and c = -1

Quadratic formula :

Substitute a = 1, b = -4 and c = -1.

When we look at the given numerical expression, it is clear that its value must be greater than 4. 

Therefore the value of the given numerical expression is

2 + √5

Problem 2 :

Solve for x in the following equation : 

4x - 3 ⋅ 2x+2 + 25  =  0

Solution :

4x - 3 ⋅ 2x+2 + 25 = 0 

(2x)- 3 ⋅ 2⋅ 22 + 32 = 0 

(2x)- 3 ⋅ 2⋅ 4 + 32 = 0

(2x)- 12 ⋅ 2x + 32 = 0

Let y = 2x.

y2 - 12y + 32 = 0 

(y - 8)(y - 4) = 0 

y - 8 = 0  or  y - 4 = 0 

y = 8  or  y = 4

Replace y by 2x.

2x = 8  or  2x = 4 

2= 23  or  2x = 22 

x = 3  or  x = 2

So, the values of x are 3 and 2.

Problem 3 :

If the sum of the roots of the quadratic equation     

ax2 + bx + c = 0

is equal to the sum of the squares of their reciprocals, then find the value  of

(b2/ac) + (bc/a2)

Solution :

Let 'α' and 'β' be the roots of the equation

ax2 + bx + c = 0

Given : Sum of the roots is equal to the sum of the squares of their reciprocals . 

α + β = (1/α)2 + (1/β)2

α + β = 1/α+ 1/β2 

α + β = (α+ β2)/(α2β2)

α + β = [(α + β)2 - 2αβ]/(αβ)2 ----(1)

In the quadratic equation ax2 + bx + c = 0,

sum of the roots = -b/a

product of the roots = c/a

α + β = -b/a

αβ = c/a 

(1)----> -b/a = [(-b/a)2 - 2c/a]/(c/a)2 

(-b/a) ⋅ (c2/a2) = b2/a2 - 2c/a

-bc2/a= b2/a2 - 2ac/a2 

-bc2/a3 = (b2 - 2ac)/a2 

-bc2 = a⋅ [(b2 - 2ac)/a2]

-bc2 = a(b2 - 2ac)

-bc2 = ab2 - 2a2c

2a2c = ab2 + bc2

Divide both sides by a2c.

2 = b2/ac + bc/a2

So, the value of (b2/ac) + (bc/a2) is 2.

Problem 4 :

If L + M + N = 0 and L, M, N are rationales, examine the nature of the roots of the equation

(M + N - L)x2 + (N + L - M)x + (L + M - N)  =  0

Solution :

Given : L + M + N = 0.

L + M = -N

M + N = -L

N + L = -M 

Given : (M + N - L)x2 + (N + L - M)x + (L + M - N) = 0.

(-L - L)x2 + (-M - M)x + (-N - N) = 0 

- 2Lx2 - 2Mx - 2N = 0

Divide both sides by -2. 

Lx2 + Mx + N = 0 


In the above quadratic equation a = L, b = M and c = N.

Substitute a = L, b = M and c = N in the discriminant of the quadratic equation (b2- 4ac). 

b2- 4ac = M2 - 4LN

b2- 4ac = (-M)2 - 4LN

Substitute -M = L + N.

b2- 4ac = (L + N)2 - 4LN

b2- 4ac = L2 + N 2 + 2LN - 4LN

b2- 4ac = L2 + N 2 - 2LN

b2- 4ac = (L - N)2 

Because b2-4ac > 0 and also a perfect square, the roots are real and rational.

Problem 5 :

If p ≠ q and p2 = 5p - 6, q2 = 5q - 6, find the quadratic equation having roots p/q and q/p. 

Solution :

Given : p2 = 5p - 6 and q2 = 5q - 6.

p- 5p + 6 = 0  and  q- 5q + 6 = 0

By solving the above quadratic equations, we get

p = -2, -3

q = -2,  -3 

Because p ≠ q, we can take p = -2 and q = -3.

p/q = -2/(-3) = 2/3

q/p = 3/2

Construction of quadratic equation :

x2 - (sum of the roots)x + product of the roots = 0

Quadratic equation having roots p/q and q/p :

x2 - (p/q + q/p)x + p/q ⋅ q/p = 0

x2 - (p/q + q/p)x + 1 = 0

Substitute p/q = 3/2 and q/p = 2/3.

x2 - (3/2 + 2/3)x + 1 = 0

x2 - (13/6)x + 1 = 0

Multiply both sides by 6. 

6x2 - 13x + 6 = 0

Hence, the required quadratic equation is

6x2 - 13x + 6 = 0

Problem 6 :

If one root of the equation x- 8x + m  =  0 exceeds the other by 4, then find the value of m.

Solution :

In the given equation x2 - 8x + m = 0, constant term m is positive.

The two factors of m must satisfy the following two conditions.

(i) Sum of the two factors of m must be equal to the middle term - 8.

(ii) One root of the equation must exceed the other by 4. That is, there must be a difference of 4 between the two roots. 

The above two conditions can be met, only if the two factors of m are  

- 2 and - 6 

Then, we have

m = (- 2) ⋅ (- 6)

m = 12

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