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 ax^{2} + bx + c = 0 and x^{2} - 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} + 2^{5} = 0
Solution :
4^{x} - 3 ⋅ 2^{x+2} + 2^{5} = 0
(2^{x})^{2 }- 3 ⋅ 2^{x }⋅ 2^{2} + 32 = 0
(2^{x})^{2 }- 3 ⋅ 2^{x }⋅ 4 + 32 = 0
(2^{x})^{2 }- 12 ⋅ 2^{x} + 32 = 0
Let y = 2^{x}.
y^{2} - 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 }= 2^{3} or 2^{x} = 2^{2}
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
ax^{2} + bx + c = 0
is equal to the sum of the squares of their reciprocals, then find the value of
(b^{2}/ac) + (bc/a^{2})
Solution :
Let 'α' and 'β' be the roots of the equation
ax^{2} + 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^{2 }+ 1/β^{2}
α + β = (α^{2 }+ β^{2})/(α^{2}β^{2})
α + β = [(α + β)^{2} - 2αβ]/(αβ)^{2} ----(1)
In the quadratic equation ax^{2} + 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) ⋅ (c^{2}/a^{2}) = b^{2}/a^{2} - 2c/a
-bc^{2}/a^{3 }= b^{2}/a^{2} - 2ac/a^{2}
-bc^{2}/a^{3} = (b^{2} - 2ac)/a^{2}
-bc^{2} = a^{3 }⋅ [(b^{2} - 2ac)/a^{2}]
-bc^{2} = a(b^{2} - 2ac)
-bc^{2} = ab^{2} - 2a^{2}c
2a^{2}c = ab^{2} + bc^{2}
Divide both sides by a^{2}c.
2 = b^{2}/ac + bc/a^{2}
So, the value of (b^{2}/ac) + (bc/a^{2}) 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)x^{2} + (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)x^{2} + (N + L - M)x + (L + M - N) = 0.
(-L - L)x^{2} + (-M - M)x + (-N - N) = 0
- 2Lx^{2} - 2Mx - 2N = 0
Divide both sides by -2.
Lx^{2} + 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 (b^{2}- 4ac).
b^{2}- 4ac = M^{2} - 4LN
b^{2}- 4ac = (-M)^{2} - 4LN
Substitute -M = L + N.
b^{2}- 4ac = (L + N)^{2} - 4LN
b^{2}- 4ac = L^{2} + N ^{2} + 2LN - 4LN
b^{2}- 4ac = L^{2} + N ^{2} - 2LN
b^{2}- 4ac = (L - N)^{2}
Because b^{2}-4ac > 0 and also a perfect square, the roots are real and rational.
Problem 5 :
If p ≠ q and p^{2} = 5p - 6, q^{2} = 5q - 6, find the quadratic equation having roots p/q and q/p.
Solution :
Given : p^{2} = 5p - 6 and q^{2} = 5q - 6.
p^{2 }- 5p + 6 = 0 and q^{2 }- 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 :
x^{2} - (sum of the roots)x + product of the roots = 0
Quadratic equation having roots p/q and q/p :
x^{2} - (p/q + q/p)x + p/q ⋅ q/p = 0
x^{2} - (p/q + q/p)x + 1 = 0
Substitute p/q = 3/2 and q/p = 2/3.
x^{2} - (3/2 + 2/3)x + 1 = 0
x^{2} - (13/6)x + 1 = 0
Multiply both sides by 6.
6x^{2} - 13x + 6 = 0
Hence, the required quadratic equation is
6x^{2} - 13x + 6 = 0
Problem 6 :
If one root of the equation x^{2 }- 8x + m = 0 exceeds the other by 4, then find the value of m.
Solution :
In the given equation x^{2} - 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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