QUADRATIC EQUATIONS WORKSHEET

Problem 1 :

Solve the following quadratic equation by factoring.

x² – 5x – 24  =  0

Problem 2 :

If one root of a quadratic equation (2 + √3), then form the equation given that the roots are irrational. 

Problem 3 : 

If α and β be the roots of x² + 7x + 12  =  0, find the quadratic equation whose roots are 

(α + β)² and (α - β)²

Problem 4 :

A piece of iron rod costs $60. If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. What is the length of the rod? 

Problem 5 : 

Find the value : 

Problem 6 :

Solve for x in the following equation : 

4^x - 3 ⋅ 2^x+2 + 2⁵  =  0

Problem 7 :

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 8 :

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

Problem 9 :

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

Problem 10 :

If one root of the equation x² - 8x + m  =  0 exceeds the other by 4, then find the value of m.    

1. Solution :

x² – 5x – 24 = 0

In the given quadratic equation, the coefficient of x² is 1.

Decompose the constant term -24 into two factors such that the product of the two factors is equal to -24 and the addition of two factors is equal to the coefficient of x, that is 5. 

Then, the two factors of -24 are

+3 and -8

Factor the given quadratic equation using +3 and -8 and solve for x.

(x + 3)(x - 8) = 0

x + 3 = 0  or  x - 8 = 0

x = -3  or  x = 8

So, the solution is {-3, 8}. 

2. Solution :

(2 + √3) is an irrational number.

We already know the fact that irrational roots of a quadratic equation will occur in conjugate pairs.

That is,  if (2 + √3) is one root of a quadratic equation, then (2 - √3) will be the other root of the same equation. 

So, (2 + √3) and (2 - √3) are the roots of the required quadratic equation.

Sum of the roots is 

= (2 + √3) + (2 - √3)

= 4

Product of the roots is

= (2 + √3)(2 - √3)

= 2² - √3²

= 4 - 3

= 1

Formation of quadratic equation :

x² - (sum of the roots)x + product of the roots = 0 

x² - 4x + 1 = 0

3. Answer :

Given : α and β are the roots of x² + 7x + 12 = 0.

sum of roots = -coefficient of x/coefficient of x²

α + β = -7/1

α + β = -7

product of roots = constant term/coefficient of x²

αβ = 12/1

αβ = 12

Quadratic equation with roots (α + β)² and (α - β)² :

x² - [(α + β)² + (α - β)²]x + (α + β)²(α - β)² = 0 ----(1)

Find the values of (α + β)² and (α - β)².

(α + β)² = (-7)²

(α + β)² = 49

(α - β)² = (α + β)² - 4αβ

(α - β)² = (-7)² - 4(12)

(α - β)² = 49 - 48

(α - β)² = 1

So, the required quadratic equation is

(1)-----> x² - [49 + 1]x + 49 ⋅ 1 = 0

x² - 50x + 49 = 0

4. Solution :

Let 'x' be the length of the given rod. 

Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). 

Cost of one meter of the given rod  is

= 60/x 

Cost of one meter of the rod which is 2 meter shorter is

= 60/(x - 2) 

Given : If the rod was 2 meter shorter and each meter costs $1 more.

That is, 60/(x - 2) is $1 more than 60/x.

[60/(x - 2)] - [60/x] = 1 

Simplify.

[60x - 60(x - 2)]/[x(x - 2)] = 1

[60x - 60x + 120]/[x² - 2x] = 1

120/(x² - 2x) = 1

120 = x² - 2x

x² + 2x - 120 = 0

(x + 10)(x - 12) = 0 

x = -10  or  x = 12

Because length can not be a negative number, we can ignore - 10. 

So, the length of the given rod is 12 m.

5. Solution :

In this question, wherever you find the pattern shown below, it can be replaced by x.

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

6. 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. 

7. 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.

8. 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.

9. 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

10. 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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