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 : 

4^x - 3 ⋅ 2^x+2 + 2⁵  =  0

Solution :

4^x - 3 ⋅ 2^x+2 + 2⁵ = 0 

(2^x)² - 3 ⋅ 2^x ⋅ 2² + 32 = 0 

(2^x)² - 3 ⋅ 2^x ⋅ 4 + 32 = 0

(2^x)² - 12 ⋅ 2^x + 32 = 0

Let y = 2^x.

y² - 12y + 32 = 0 

(y - 8)(y - 4) = 0 

y - 8 = 0  or  y - 4 = 0 

y = 8  or  y = 4

Replace y by 2^x.

2^x = 8  or  2^x = 4 

2^x = 2³  or  2^x = 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/α² + 1/β² 

α + β = (α² + 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/a³ = b²/a² - 2ac/a² 

-bc²/a³ = (b² - 2ac)/a² 

-bc² = a³ ⋅ [(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.

p² - 5p + 6 = 0  and  q² - 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 x² - 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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