PROBLEMS ON QUADRATIC EQUATIONS

Problem 1 : 

Find the value : 

Solution :

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

Hence, 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.

Then, we have

y² - 12y + 32  =  0 

(y - 8) ⋅ (y - 4)  =  0 

y - 8 = 0 or y - 4  =  0 

y  =  8 or y  =  4

Plug y  =  2^x.

2^x  =  8 or 2^x  =  4 

2^x  =  2³ or 2^x  =  2² 

x  =  3 or  x  =  2

Hence, 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 . 

So, we have

α + β  =  (1 / α)²  + (1 / β)²

Simplify.

α + β  =  1/ α²  + 1 / β² 

α + β  =  (α² + β²) /  (α² ⋅ β²)

 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

So, we have

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

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

Problem 4 :

If L + M + N = 0 and L, M, N are rationals, then, 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

Then, we have

L + M  =  - N

M + N  =  - L

N + L  =  - M 

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

So, we have

(- 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.

Plug 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 

Plug -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

Then we have

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.

Then, we have

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

Plug 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

Hence, the value of m is 12.

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