GSEB Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.2

Question 1. Find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficients
1. \( x^2 - 2x - 8 \)
2. \( 4s^2 - 4s + 1 \)
3. \( 6x^2 - 7x - 3 \)
4. \( 4u^2 + 8u \)
5. \( t^2 - 15 \)
6. \( 3x^2 - x - 4 \)
Answer:
1. We have \( p(x) = x^2 - 2x - 8 \)
To find the zeroes, we set \( p(x) = 0 \).
\( x^2 - 2x - 8 = 0 \)
By splitting the middle term, we get:
\( x^2 - 4x + 2x - 8 = 0 \)
\( x(x - 4) + 2(x - 4) = 0 \)
\( (x - 4)(x + 2) = 0 \)

\( \implies x - 4 = 0 \) or \( x + 2 = 0 \)
\( \implies x = 4 \) or \( x = -2 \)
The roots of the polynomial \( p(x) \) are 4 and -2.
We can let \( \alpha = 4 \) and \( \beta = -2 \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = 4 + (-2) = 2 \)
So, \( \alpha + \beta = 2 \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } x)}{\text{coefficient of } x^2} = \frac{-b}{a} \).
From the polynomial \( x^2 - 2x - 8 \), we have \( a = 1 \), \( b = -2 \), \( c = -8 \).
So, \( \frac{-b}{a} = \frac{-(-2)}{1} = 2 \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = 4 \times (-2) = -8 \)
So, \( \alpha\beta = -8 \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{-8}{1} = -8 \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: First, we find the numbers that make the polynomial zero. These are 4 and -2. Then, we check if adding these numbers and multiplying them matches what the formula for polynomial coefficients predicts. Both the sum and product match, showing the connection is correct.

Exam Tip: Remember to clearly state the values of a, b, and c from the polynomial before applying the sum and product formulas for verification.

Answer:
2. We have \( p(s) = 4s^2 - 4s + 1 \)
To find the zeroes, we set \( p(s) = 0 \).
\( 4s^2 - 4s + 1 = 0 \)
By splitting the middle term, we get:
\( 4s^2 - 2s - 2s + 1 = 0 \)
\( 2s(2s - 1) - 1(2s - 1) = 0 \)
\( (2s - 1)(2s - 1) = 0 \)

\( \implies 2s - 1 = 0 \) or \( 2s - 1 = 0 \)
\( \implies 2s = 1 \) or \( 2s = 1 \)
\( \implies s = \frac{1}{2} \) or \( s = \frac{1}{2} \)
The roots of the polynomial \( p(s) \) are \( \frac{1}{2} \) and \( \frac{1}{2} \).
We can let \( \alpha = \frac{1}{2} \) and \( \beta = \frac{1}{2} \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = \frac{1}{2} + \frac{1}{2} = 1 \)
So, \( \alpha + \beta = 1 \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } s)}{\text{coefficient of } s^2} = \frac{-b}{a} \).
From the polynomial \( 4s^2 - 4s + 1 \), we have \( a = 4 \), \( b = -4 \), \( c = 1 \).
So, \( \frac{-b}{a} = \frac{-(-4)}{4} = \frac{4}{4} = 1 \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
So, \( \alpha\beta = \frac{1}{4} \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } s^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{1}{4} \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: First, we find the values of 's' that make the polynomial zero, which are both \( \frac{1}{2} \). Then, we check if adding and multiplying these values matches the rules for polynomial coefficients. Both the sum and the product are verified to be correct.

Exam Tip: Be careful with signs when substituting values into \( \frac{-b}{a} \), especially when 'b' itself is negative.

Answer:
3. We have \( p(x) = 6x^2 - 3 - 7x \). We can rewrite this as \( p(x) = 6x^2 - 7x - 3 \).
To find the zeroes, we set \( p(x) = 0 \).
\( 6x^2 - 7x - 3 = 0 \)
By splitting the middle term, we get:
\( 6x^2 - 9x + 2x - 3 = 0 \)
\( 3x(2x - 3) + 1(2x - 3) = 0 \)
\( (2x - 3)(3x + 1) = 0 \)

\( \implies 2x - 3 = 0 \) or \( 3x + 1 = 0 \)
\( \implies 2x = 3 \) or \( 3x = -1 \)
\( \implies x = \frac{3}{2} \) or \( x = \frac{-1}{3} \)
The roots of the polynomial \( p(x) \) are \( \frac{3}{2} \) and \( \frac{-1}{3} \).
We can let \( \alpha = \frac{3}{2} \) and \( \beta = \frac{-1}{3} \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = \frac{3}{2} + (\frac{-1}{3}) = \frac{9 - 2}{6} = \frac{7}{6} \)
So, \( \alpha + \beta = \frac{7}{6} \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } x)}{\text{coefficient of } x^2} = \frac{-b}{a} \).
From the polynomial \( 6x^2 - 7x - 3 \), we have \( a = 6 \), \( b = -7 \), \( c = -3 \).
So, \( \frac{-b}{a} = \frac{-(-7)}{6} = \frac{7}{6} \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = \frac{3}{2} \times (\frac{-1}{3}) = \frac{-3}{6} = \frac{-1}{2} \)
So, \( \alpha\beta = \frac{-1}{2} \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{-3}{6} = \frac{-1}{2} \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: First, we set the polynomial equal to zero and solve it to find its roots, which are \( \frac{3}{2} \) and \( \frac{-1}{3} \). Then, we check if the sum and product of these roots match the values calculated from the polynomial's coefficients. Both relationships hold true.

Exam Tip: Always rearrange the polynomial into standard form \( ax^2 + bx + c \) before identifying the coefficients a, b, and c to avoid errors.

Answer:
4. We have \( p(u) = 4u^2 + 8u \).
To find the zeroes, we set \( p(u) = 0 \).
\( 4u^2 + 8u = 0 \)
We can factor out \( 4u \):
\( 4u(u + 2) = 0 \)

\( \implies 4u = 0 \) or \( u + 2 = 0 \)
\( \implies u = 0 \) or \( u = -2 \)
The roots of the polynomial \( p(u) \) are 0 and -2.
We can let \( \alpha = 0 \) and \( \beta = -2 \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = 0 + (-2) = -2 \)
So, \( \alpha + \beta = -2 \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } u)}{\text{coefficient of } u^2} = \frac{-b}{a} \).
From the polynomial \( 4u^2 + 8u \), we have \( a = 4 \), \( b = 8 \), \( c = 0 \) (since there is no constant term).
So, \( \frac{-b}{a} = \frac{-8}{4} = -2 \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = 0 \times (-2) = 0 \)
So, \( \alpha\beta = 0 \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } u^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{0}{4} = 0 \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: First, we find the roots of the polynomial by factoring it, which are 0 and -2. Then, we check if the sum and product of these roots match the values derived from the polynomial's coefficients, showing the verification.

Exam Tip: Remember that if a term is missing in a polynomial, its coefficient is 0. For example, in \( 4u^2 + 8u \), the constant term \( c \) is 0.

Answer:
5. We have \( p(t) = t^2 - 15 \).
To find the zeroes, we set \( p(t) = 0 \).
\( t^2 - 15 = 0 \)
We can rewrite this using the difference of squares formula \( (a^2 - b^2 = (a-b)(a+b)) \):
\( t^2 - (\sqrt{15})^2 = 0 \)
\( (t - \sqrt{15})(t + \sqrt{15}) = 0 \)

\( \implies t - \sqrt{15} = 0 \) or \( t + \sqrt{15} = 0 \)
\( \implies t = \sqrt{15} \) or \( t = -\sqrt{15} \)
The roots of the polynomial \( p(t) \) are \( \sqrt{15} \) and \( -\sqrt{15} \).
We can let \( \alpha = \sqrt{15} \) and \( \beta = -\sqrt{15} \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = \sqrt{15} + (-\sqrt{15}) = 0 \)
So, \( \alpha + \beta = 0 \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } t)}{\text{coefficient of } t^2} = \frac{-b}{a} \).
From the polynomial \( t^2 - 15 \), we have \( a = 1 \), \( b = 0 \) (since there is no t-term), \( c = -15 \).
So, \( \frac{-b}{a} = \frac{-0}{1} = 0 \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = \sqrt{15} \times (-\sqrt{15}) = -15 \)
So, \( \alpha\beta = -15 \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } t^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{-15}{1} = -15 \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: We find the roots of the polynomial \( t^2 - 15 \) by using the difference of squares, which gives \( \sqrt{15} \) and \( -\sqrt{15} \). Then, we check if the sum and product of these roots match the expected values from the polynomial's coefficients, confirming the relationship.

Exam Tip: Polynomials of the form \( x^2 - k \) can always be factored using the difference of squares, where \( k \) is treated as \( (\sqrt{k})^2 \).

Answer:
6. We have \( p(x) = 3x^2 - x - 4 \).
To find the zeroes, we set \( p(x) = 0 \).
\( 3x^2 - x - 4 = 0 \)
By splitting the middle term, we get:
\( 3x^2 - 4x + 3x - 4 = 0 \)
\( x(3x - 4) + 1(3x - 4) = 0 \)
\( (3x - 4)(x + 1) = 0 \)

\( \implies 3x - 4 = 0 \) or \( x + 1 = 0 \)
\( \implies 3x = 4 \) or \( x = -1 \)
\( \implies x = \frac{4}{3} \) or \( x = -1 \)
The roots of the polynomial \( p(x) \) are \( \frac{4}{3} \) and -1.
We can let \( \alpha = \frac{4}{3} \) and \( \beta = -1 \).

Verification of relationship between roots and coefficients:
Sum of the roots \( = \frac{4}{3} + (-1) = \frac{4 - 3}{3} = \frac{1}{3} \)
So, \( \alpha + \beta = \frac{1}{3} \).
We also know that \( \text{Sum of zeroes} = \frac{-(\text{coefficient of } x)}{\text{coefficient of } x^2} = \frac{-b}{a} \).
From the polynomial \( 3x^2 - x - 4 \), we have \( a = 3 \), \( b = -1 \), \( c = -4 \).
So, \( \frac{-b}{a} = \frac{-(-1)}{3} = \frac{1}{3} \).
This confirms that \( \alpha + \beta = \frac{-b}{a} \).

Product of the roots \( = \frac{4}{3} \times (-1) = \frac{-4}{3} \)
So, \( \alpha\beta = \frac{-4}{3} \).
We also know that \( \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a} \).
So, \( \frac{c}{a} = \frac{-4}{3} \).
This confirms that \( \alpha\beta = \frac{c}{a} \).
In simple words: First, we find the values of 'x' that make the polynomial zero, which are \( \frac{4}{3} \) and -1. Then, we check if adding these numbers and multiplying them matches what the formulas for polynomial coefficients predict. Both the sum and product relationships are verified as correct.

Exam Tip: When dealing with fractional roots, ensure you correctly perform addition and multiplication of fractions, paying attention to common denominators for addition.

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
1. \( \frac{1}{4}, -1 \)
2. \( \sqrt{2}, \frac{1}{3} \)
3. \( 0, \sqrt{5} \)
4. \( 1, 1 \)
5. \( \frac{-1}{4}, \frac{1}{4} \)
6. \( 4, 1 \)
Answer:
We know that a quadratic polynomial, when the sum (S) and product (P) of its zeroes are given, is \( p(x) = K[x^2 - Sx + P] \), where K is a constant.

1. Given: Sum of the zeroes \( S = \frac{1}{4} \), Product of the zeroes \( P = -1 \).
The required polynomial is:
\( p(x) = K[x^2 - (\frac{1}{4})x + (-1)] \)
\( p(x) = K[x^2 - \frac{1}{4}x - 1] \)
To remove the fraction, we can set \( K = 4 \):
\( p(x) = 4[x^2 - \frac{1}{4}x - 1] \)
\( p(x) = 4x^2 - x - 4 \)
So, a quadratic polynomial is \( 4x^2 - x - 4 \).
In simple words: We use the formula \( K[x^2 - Sx + P] \) with the given sum and product. Then, we choose a value for K to remove any fractions, giving us the polynomial \( 4x^2 - x - 4 \).

Exam Tip: Always remember the general form of a quadratic polynomial in terms of its sum (S) and product (P) of zeroes: \( K(x^2 - Sx + P) \).

Answer:
2. Given: Sum of the zeroes \( S = \sqrt{2} \), Product of the zeroes \( P = \frac{1}{3} \).
The required polynomial is:
\( p(x) = K[x^2 - (\sqrt{2})x + \frac{1}{3}] \)
\( p(x) = K[x^2 - \sqrt{2}x + \frac{1}{3}] \)
To remove the fraction, we can set \( K = 3 \):
\( p(x) = 3[x^2 - \sqrt{2}x + \frac{1}{3}] \)
\( p(x) = 3x^2 - 3\sqrt{2}x + 1 \)
So, a quadratic polynomial is \( 3x^2 - 3\sqrt{2}x + 1 \).
In simple words: We place the given sum \( \sqrt{2} \) and product \( \frac{1}{3} \) into the standard polynomial formula. To get rid of the fraction, we multiply everything by 3, resulting in the polynomial \( 3x^2 - 3\sqrt{2}x + 1 \).

Exam Tip: When removing fractions, choose K as the least common multiple of the denominators present in the polynomial expression.

Answer:
3. Given: Sum of the zeroes \( S = 0 \), Product of the zeroes \( P = \sqrt{5} \).
The required polynomial is:
\( p(x) = K[x^2 - (0)x + \sqrt{5}] \)
\( p(x) = K[x^2 - 0x + \sqrt{5}] \)
\( p(x) = K[x^2 + \sqrt{5}] \)
We can simply set \( K = 1 \):
\( p(x) = x^2 + \sqrt{5} \)
So, a quadratic polynomial is \( x^2 + \sqrt{5} \).
In simple words: Using the given sum of 0 and product of \( \sqrt{5} \) in the polynomial formula, we simplify it to \( K[x^2 + \sqrt{5}] \). By choosing K as 1, the polynomial becomes \( x^2 + \sqrt{5} \).

Exam Tip: If the sum of zeroes is 0, the \( x \) term in the quadratic polynomial will vanish, leaving only \( x^2 \) and the constant term.

Answer:
4. Given: Sum of the zeroes \( S = 1 \), Product of the zeroes \( P = 1 \).
The required polynomial is:
\( p(x) = K[x^2 - (1)x + 1] \)
\( p(x) = K[x^2 - x + 1] \)
We can simply set \( K = 1 \):
\( p(x) = x^2 - x + 1 \)
So, a quadratic polynomial is \( x^2 - x + 1 \).
In simple words: We put the given sum (1) and product (1) into the standard polynomial formula. By choosing K as 1, we get the quadratic polynomial \( x^2 - x + 1 \).

Exam Tip: When no fractions are present, it is often simplest to assume \( K = 1 \) for the required polynomial, unless specified otherwise.

Answer:
5. Given: Sum of the zeroes \( S = \frac{-1}{4} \), Product of the zeroes \( P = \frac{1}{4} \).
The required polynomial is:
\( p(x) = K[x^2 - (\frac{-1}{4})x + \frac{1}{4}] \)
\( p(x) = K[x^2 + \frac{1}{4}x + \frac{1}{4}] \)
To remove the fractions, we can set \( K = 4 \):
\( p(x) = 4[x^2 + \frac{1}{4}x + \frac{1}{4}] \)
\( p(x) = 4x^2 + x + 1 \)
So, a quadratic polynomial is \( 4x^2 + x + 1 \).
In simple words: We use the given sum \( \frac{-1}{4} \) and product \( \frac{1}{4} \) in the standard polynomial formula. To eliminate the fractions, we multiply by 4, which gives us the polynomial \( 4x^2 + x + 1 \).

Exam Tip: Be cautious with double negative signs when substituting the sum of zeroes if S is negative, i.e., \( -(-S) = +S \).

Answer:
6. Given: Sum of the zeroes \( S = 4 \), Product of the zeroes \( P = 1 \).
The required polynomial is:
\( p(x) = K[x^2 - (4)x + 1] \)
\( p(x) = K[x^2 - 4x + 1] \)
We can simply set \( K = 1 \):
\( p(x) = x^2 - 4x + 1 \)
So, a quadratic polynomial is \( x^2 - 4x + 1 \).
In simple words: We place the given sum (4) and product (1) into the standard polynomial formula. By choosing K as 1, the quadratic polynomial is \( x^2 - 4x + 1 \).

Exam Tip: If the problem does not specify a value for K, assuming \( K = 1 \) is standard practice after clearing any common denominators.