RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.3

Question 1. Test, whether the following equation are quadratic equation.
(i) \( x(x + 1) + 8 = (x + 2)(x - 2) \)
(ii) \( (x + 2)^3 = x^3 - 4 \)
(iii) \( x^2 + 3x + 1 = (x - 2)^2 \)
(iv) \( x + \frac { 1 }{ x } + x^2 = 0, x \neq 0 \)
Answer:
(i) For the equation \( x(x + 1) + 8 = (x + 2)(x - 2) \):
First, expand both sides.
\( x^2 + x + 8 = x^2 - 4 \)
Next, move all terms to one side of the equation.
\( x^2 + x + 8 - x^2 + 4 = 0 \)
Combine the like terms.
\( x + 12 = 0 \)
Since the highest power of \( x \) in the simplified equation is 1, this is a linear equation. Therefore, it is not a quadratic equation. This simple algebraic reduction helps classify the equation.
(ii) For the equation \( (x + 2)^3 = x^3 - 4 \):
First, expand the left side using the formula \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \).
\( x^3 + 2^3 + 3 \cdot x \cdot 2 (x + 2) = x^3 - 4 \)
\( x^3 + 8 + 6x(x + 2) = x^3 - 4 \)
\( x^3 + 8 + 6x^2 + 12x = x^3 - 4 \)
Next, move all terms to one side of the equation.
\( x^3 + 6x^2 + 12x + 8 - x^3 + 4 = 0 \)
Combine the like terms.
\( 6x^2 + 12x + 12 = 0 \)
Since the highest power of \( x \) in the simplified equation is 2, this is a quadratic equation.
(iii) For the equation \( x^2 + 3x + 1 = (x - 2)^2 \):
First, expand the right side using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \).
\( x^2 + 3x + 1 = x^2 - 4x + 4 \)
Next, move all terms to one side of the equation.
\( x^2 + 3x + 1 - x^2 + 4x - 4 = 0 \)
Combine the like terms.
\( 7x - 3 = 0 \)
Since the highest power of \( x \) in the simplified equation is 1, this is a linear equation. Therefore, it is not a quadratic equation. Quadratic equations always have an \( x^2 \) term.
(iv) For the equation \( x + \frac { 1 }{ x } + x^2 = 0, x \neq 0 \):
To simplify, multiply the entire equation by \( x \) to remove the fraction.
\( x \cdot x + \frac { 1 }{ x } \cdot x + x^2 \cdot x = 0 \cdot x \)
\( x^2 + 1 + x^3 = 0 \)
Rearrange the terms in descending order of power.
\( x^3 + x^2 + 1 = 0 \)
Since the highest power of \( x \) in the simplified equation is 3, this is a cubic equation. Therefore, it is not a quadratic equation. This type of simplification is crucial in solving rational equations.
In simple words: To check if an equation is quadratic, simplify it as much as possible. If the biggest power of the variable (like \( x \)) is 2, then it is quadratic. If it's 1, it's linear. If it's 3, it's cubic.

🎯 Exam Tip: Always fully simplify both sides of the equation before determining its type. An equation might look quadratic initially but simplify to a linear one, or vice-versa.

Question 2. Solve the following equation by factorization.
(ii) \( 9x^2 - 3x - 2 = 0 \)
(iii) \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \)
(iv) \( x^2 - 8x + 16 = 0 \)
(v) \( \frac { 1 }{ x-2 } + \frac { 2 }{ x-1 } = \frac { 6 }{ x } \) where \( x \neq 1, 2 \)
(vi) \( 100x^2 - 20x + 1 = 0 \)
(viii) \( x^2 + 8x + 7 = 0 \)
(x) \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
(xi) \( abx^2 + (b^2 - ac)x - bc = 0 \)
Answer:
(ii) For the equation \( 9x^2 - 3x - 2 = 0 \):
To factorize, split the middle term \( -3x \) into \( -6x + 3x \).
\( 9x^2 - 6x + 3x - 2 = 0 \)
Group the terms and factor out common factors.
\( 3x(3x - 2) + 1(3x - 2) = 0 \)
Factor out the common binomial factor \( (3x - 2) \).
\( (3x - 2)(3x + 1) = 0 \)
Set each factor to zero to find the roots.
\( 3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3} \)
\( 3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3} \)
The roots of the equation are \( \frac{2}{3} \) and \( -\frac{1}{3} \).
(iii) For the equation \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \):
To factorize, split the middle term \( 10x \) into \( 3x + 7x \).
\( \sqrt{3}x^2 + 3x + 7x + 7\sqrt{3} = 0 \)
Group the terms and factor out common factors.
\( \sqrt{3}x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0 \)
Factor out the common binomial factor \( (x + \sqrt{3}) \).
\( (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \)
Set the first factor to zero to find one root.
\( x + \sqrt{3} = 0 \implies x = -\sqrt{3} \)
(Note: The source did not provide the solution for the second root.)
(iv) For the equation \( x^2 - 8x + 16 = 0 \):
This is a perfect square trinomial. Split the middle term \( -8x \) into \( -4x - 4x \).
\( x^2 - 4x - 4x + 16 = 0 \)
Group the terms and factor out common factors.
\( x(x - 4) - 4(x - 4) = 0 \)
Factor out the common binomial factor \( (x - 4) \).
\( (x - 4)(x - 4) = 0 \)
Set the factor to zero to find the repeated root.
\( x - 4 = 0 \implies x = 4 \)
The roots of the equation are 4 and 4.
(v) For the equation \( \frac { 1 }{ x-2 } + \frac { 2 }{ x-1 } = \frac { 6 }{ x } \):
First, combine the fractions on the left-hand side by finding a common denominator.
\( \frac { (x-1) + 2(x-2) }{ (x-2)(x-1) } = \frac { 6 }{ x } \)
\( \frac { x-1 + 2x-4 }{ x^2-3x+2 } = \frac { 6 }{ x } \)
\( \frac { 3x-5 }{ x^2-3x+2 } = \frac { 6 }{ x } \)
Next, cross-multiply to eliminate the denominators.
\( x(3x-5) = 6(x^2-3x+2) \)
\( 3x^2 - 5x = 6x^2 - 18x + 12 \)
Move all terms to one side to form a standard quadratic equation.
\( 0 = 6x^2 - 3x^2 - 18x + 5x + 12 \)
\( 3x^2 - 13x + 12 = 0 \)
To factorize, split the middle term \( -13x \) into \( -4x - 9x \).
\( 3x^2 - 4x - 9x + 12 = 0 \)
Group the terms and factor out common factors.
\( x(3x - 4) - 3(3x - 4) = 0 \)
Factor out the common binomial factor \( (3x - 4) \).
\( (x - 3)(3x - 4) = 0 \)
Set each factor to zero to find the roots.
\( x - 3 = 0 \implies x = 3 \)
\( 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \)
The roots of the equation are 3 and \( \frac{4}{3} \). This type of problem often requires careful algebraic manipulation before factorization.
(vi) For the equation \( 100x^2 - 20x + 1 = 0 \):
This equation is a perfect square trinomial. Split the middle term \( -20x \) into \( -10x - 10x \).
\( 100x^2 - 10x - 10x + 1 = 0 \)
Group the terms and factor out common factors.
\( 10x(10x - 1) - 1(10x - 1) = 0 \)
Factor out the common binomial factor \( (10x - 1) \).
\( (10x - 1)(10x - 1) = 0 \)
Set the factor to zero to find the repeated root.
\( 10x - 1 = 0 \implies 10x = 1 \implies x = \frac{1}{10} \)
The roots of the equation are \( \frac{1}{10} \) and \( \frac{1}{10} \).
(viii) For the equation \( x^2 + 8x + 7 = 0 \):
To factorize, split the middle term \( 8x \) into \( x + 7x \).
\( x^2 + x + 7x + 7 = 0 \)
Group the terms and factor out common factors.
\( x(x + 1) + 7(x + 1) = 0 \)
Factor out the common binomial factor \( (x + 1) \).
\( (x + 1)(x + 7) = 0 \)
Set each factor to zero to find the roots.
\( x + 1 = 0 \implies x = -1 \)
\( x + 7 = 0 \implies x = -7 \)
The roots of the equation are -1 and -7.
(x) For the equation \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \):
First, factor the constant term \( (a^4 - b^4) \) as a difference of squares: \( (a^2+b^2)(a^2-b^2) \).
Next, split the middle term \( -4a^2x \) into two parts that will help in factorization: \( -2(a^2+b^2)x \) and \( -2(a^2-b^2)x \).
\( 4x^2 - 2(a^2+b^2)x - 2(a^2-b^2)x + (a^2+b^2)(a^2-b^2) = 0 \)
Group the terms and factor out common factors.
\( 2x[2x - (a^2+b^2)] - (a^2-b^2)[2x - (a^2+b^2)] = 0 \)
Factor out the common binomial factor \( [2x - (a^2+b^2)] \).
\( [2x - (a^2+b^2)][2x - (a^2-b^2)] = 0 \)
Set each factor to zero to find the roots.
\( 2x - (a^2+b^2) = 0 \implies 2x = a^2+b^2 \implies x = \frac{a^2+b^2}{2} \)
\( 2x - (a^2-b^2) = 0 \implies 2x = a^2-b^2 \implies x = \frac{a^2-b^2}{2} \)
The roots of the equation are \( \frac{a^2+b^2}{2} \) and \( \frac{a^2-b^2}{2} \). This complex factorization demonstrates how to solve quadratics with literal coefficients.
(xi) For the equation \( abx^2 + (b^2 - ac)x - bc = 0 \):
To factorize, split the middle term \( (b^2 - ac)x \) into \( b^2x - acx \).
\( abx^2 + b^2x - acx - bc = 0 \)
Group the terms and factor out common factors.
\( bx(ax + b) - c(ax + b) = 0 \)
Factor out the common binomial factor \( (ax + b) \).
\( (ax + b)(bx - c) = 0 \)
Set each factor to zero to find the roots.
\( ax + b = 0 \implies ax = -b \implies x = -\frac{b}{a} \)
\( bx - c = 0 \implies bx = c \implies x = \frac{c}{b} \)
The roots of the equation are \( -\frac{b}{a} \) and \( \frac{c}{b} \).
In simple words: To solve a quadratic equation by factorization, you usually split the middle term into two parts. Then, group the terms and take out common factors. This process leads to two factors which, when set to zero, give you the two solutions or roots of the equation.

🎯 Exam Tip: When factorizing, remember to check your work by multiplying the factors back together to ensure they equal the original quadratic expression. This is especially important for complex expressions or those with literal coefficients.