**Quadratic Equations Worksheet :**

Worksheet given in this section would be much useful to the students who would like to practice problems on quadratic equations.

**Problem 1 : **

Find the value :

**Problem 2 :**

Solve for x in the following equation :

4^{x} - 3 ⋅ 2^{x+2} + 2^{5} = 0

**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)

**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

**Problem 5 :**

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

**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

Let us look at the next problem on "Quadratic equations worksheet".

**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}.

Then, we have

y^{2} - 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} = 2^{3} or 2^{x} = 2^{2}

x = 3 or x = 2

Hence, the values of x are 3 and 2.

Let us look at the next problem on "Quadratic equations worksheet".

**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

ax^{2} + 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/ 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 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) ⋅ (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}

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

Let us look at the next problem on "Quadratic equations worksheet".

**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)x^{2} + (N + L - M)x + (L + M - N) = 0

So, we have

(- 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.

Plug 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

Plug -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.

Let us look at the next problem on "Quadratic equations worksheet".

**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 :** p^{2} = 5p - 6 and q^{2} = 5q - 6

Then we have

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.

Then, we have

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

Plug 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

Let us look at the next problem on "Quadratic equations worksheet".

**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 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

Hence, the value of m is 12.

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