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 :

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.

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  =  2x.

2x  =  8 or 2x  =  4

2x  =  23 or 2x  =  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  =  (α+ β2) /  (α⋅ 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

-bc/ a3  =  b2/a2 - 2ac/a2

-bc/ 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

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

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.

Then, we have

p/q  =  -2 / -3  =  2/3

q/p  =  3/2

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 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

Hence, the value of m is 12. Apart from the problems given above, if you need more problems on quadratic equations

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