CBSE Class 10 Mathematics Polynomials Assignment Set 05

Polynomial:
An expression of the form \(p(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n\) where \(a_n \neq 0\) is called a polynomial in variable \(x\) of degree \(n\), where \(a_0, a_1, \dots, a_n\) are real numbers and each power of \(x\) is a non-negative integer.
Example: \(2x^2 - 5x + 1\) is a polynomial of degree 2.
Note: \(\sqrt{x} + 3\) is not a polynomial.

Types of polynomials based on degree:
- A polynomial \(p(x) = ax + b\) of degree 1 is called a linear polynomial. Examples: \(5x - 3\), \(2x\), etc.
- A polynomial \(p(x) = ax^2 + bx + c\) of degree 2 is called a quadratic polynomial. Examples: \(2x^2 + x - 1\), \(1 - 5x + x^2\), etc.
- A polynomial \(p(x) = ax^3 + bx^2 + cx + d\) of degree 3 is called a cubic polynomial. Examples: \(\sqrt{3}x^3 - x + \sqrt{5}\), \(x^3 - 1\), etc.

Zeroes of a polynomial:
A real number \(k\) is called a zero of polynomial \(p(x)\) if \(p(k) = 0\). The graph of \(y = p(x)\) intersects the \(X\)-axis at points whose \(x\)-coordinates are the zeroes of the polynomial.
- A linear polynomial has only 1 zero.
- A quadratic polynomial has at most 2 zeroes.
- A cubic polynomial has at most 3 zeroes.

Relationship between zeroes and coefficients of a quadratic polynomial:
If \(\alpha, \beta\) are zeroes of \(P(x) = ax^2 + bx + c\), then:
1. Sum of zeroes = \(\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\)
2. Product of zeroes = \(\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\)

A quadratic polynomial whose zeroes are \(\alpha\) and \(\beta\) is given by:
\(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - (\text{sum of zeroes})x + \text{product of zeroes}\)

Relationship between zeroes and coefficients of a cubic polynomial:
If \(\alpha, \beta, \gamma\) are zeroes of the cubic polynomial \(p(x) = ax^3 + bx^2 + cx + d\), then:
- \(\alpha + \beta + \gamma = -\frac{b}{a}\)
- \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\)
- \(\alpha\beta\gamma = -\frac{d}{a}\)

Division algorithm for polynomials:
If \(p(x)\) and \(g(x)\) are any two polynomials with \(g(x) \neq 0\), then we can find polynomials \(q(x)\) and \(r(x)\) such that:
\(p(x) = q(x) \cdot g(x) + r(x)\), where \(r(x) = 0\) or degree of \(r(x) <\) degree of \(g(x)\).

( Level - 1 )

Question 1. In a graph of \(y = p(x)\), find the number of zeroes of \(p(x)\).
Answer: 3
In simple words: The graph intersects the X-axis at 3 distinct points, so the polynomial has 3 zeroes.

Exam Tip: Count the exact number of times the curve intersects or touches the horizontal X-axis.

Question 2. If \(\alpha, \beta\) are the zeroes of \(f(x) = x^2 + x + 1\), then find \(\frac{1}{\alpha} + \frac{1}{\beta}\).
Answer:
For the polynomial \(x^2 + x + 1\), comparing with \(ax^2 + bx + c\), we get \(a = 1, b = 1, c = 1\).
Sum of zeroes: \(\alpha + \beta = -\frac{b}{a} = \frac{-1}{1} = -1\)
Product of zeroes: \(\alpha\beta = \frac{c}{a} = \frac{1}{1} = 1\)
Now, \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{1} = -1\)
In simple words: Take the LCM of the denominators to express the expression in terms of sum and product of zeroes.

Exam Tip: Always convert algebraic expressions of zeroes into standard forms of \((\alpha + \beta)\) and \(\alpha\beta\).

Question 3. Find a quadratic polynomial whose zeroes are \(-\frac{2}{\sqrt{3}}\) and \(\frac{\sqrt{3}}{4}\).
Answer:
Let \(\alpha = -\frac{2}{\sqrt{3}}\) and \(\beta = \frac{\sqrt{3}}{4}\).
Sum of zeroes: \(\alpha + \beta = -\frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{4} = \frac{-8 + 3}{4\sqrt{3}} = -\frac{5}{4\sqrt{3}}\)
Product of zeroes: \(\alpha\beta = \left(-\frac{2}{\sqrt{3}}\right)\left(\frac{\sqrt{3}}{4}\right) = -\frac{2}{4} = -\frac{1}{2}\)
The required quadratic polynomial is:
\(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
\(p(x) = x^2 - \left(-\frac{5}{4\sqrt{3}}\right)x + \left(-\frac{1}{2}\right)\)
\(p(x) = x^2 + \frac{5}{4\sqrt{3}}x - \frac{1}{2}\)
In simple words: Calculate the sum and product of the given numbers, then substitute them into the standard quadratic formula.

Exam Tip: You may also multiply the entire polynomial by a constant to clear fractions if required by options.

Question 4. If \(p(x) = \frac{1}{3}x^2 - 5x + \frac{3}{2}\), then find its sum and product of zeroes.
Answer:
Comparing \(p(x) = \frac{1}{3}x^2 - 5x + \frac{3}{2}\) with \(ax^2 + bx + c\):
\(a = \frac{1}{3}, \quad b = -5, \quad c = \frac{3}{2}\)
Sum of zeroes = \(-\frac{b}{a} = -\frac{-5}{\frac{1}{3}} = 5 \times 3 = 15\)
Product of zeroes = \(\frac{c}{a} = \frac{\frac{3}{2}}{\frac{1}{3}} = \frac{3}{2} \times 3 = \frac{9}{2}\)
In simple words: Identify coefficients \(a, b, c\) correctly even when they are fractions, then apply formulas.

Exam Tip: Be careful with complex fractions when dividing by \(a\) when \(a\) is a fraction.

Question 5. If the sum of zeroes of a given polynomial \(f(x) = x^3 - 3kx^2 - x + 30\) is 6, find the value of \(k\).
Answer:
For the cubic polynomial \(f(x) = x^3 - 3kx^2 - x + 30\), comparing with \(ax^3 + bx^2 + cx + d\):
\(a = 1, \quad b = -3k, \quad c = -1, \quad d = 30\)
Sum of zeroes = \(\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-3k}{1} = 3k\)
Given that sum of zeroes = 6.
Therefore, \(3k = 6\)
\(k = 2\)
In simple words: Use the relationship between the sum of zeroes and the cubic coefficients to set up a simple linear equation in \(k\).

Exam Tip: The sum of zeroes for a cubic polynomial \(ax^3 + bx^2 + cx + d\) is always given by \(-b/a\).

Question 6. Find the zero of polynomial \(3x + 4\).
Answer:
To find the zero of the polynomial, set \(p(x) = 0\):
\(3x + 4 = 0\)
\(3x = -4\)
\(x = -\frac{4}{3}\)
In simple words: Equate the linear expression to zero and solve for \(x\).

Exam Tip: The zero of a linear polynomial \(ax + b\) is always given by \(-b/a\).

Question 7. Write the degree of zero polynomial.
Answer: Not defined.
In simple words: The zero polynomial has no non-zero terms, so its degree cannot be determined.

Exam Tip: Remember that the degree of the zero polynomial \(P(x) = 0\) is undefined, whereas the degree of a non-zero constant polynomial is 0.

( Level - 2 )

Question 1. Form a cubic polynomial with zeroes 3, 2 and -1.
Answer:
Let the zeroes be \(\alpha = 3, \beta = 2, \gamma = -1\).
Sum of zeroes: \(\alpha + \beta + \gamma = 3 + 2 + (-1) = 4\)
Sum of product of zeroes taken two at a time:
\(\alpha\beta + \beta\gamma + \gamma\alpha = (3)(2) + (2)(-1) + (-1)(3) = 6 - 2 - 3 = 1\)
Product of zeroes: \(\alpha\beta\gamma = (3)(2)(-1) = -6\)
The required cubic polynomial is given by:
\(p(x) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma\)
\(p(x) = x^3 - 4x^2 + x + 6\)
In simple words: Calculate the three fundamental symmetric sums of the zeroes and substitute them into the standard cubic formula.

Exam Tip: Pay close attention to alternating signs in the cubic polynomial formula: \((+, -, +, -)\).

Question 2. Find the zeroes of the quadratic polynomial \(6x^2 - 3 - 7x\) and verify the relationship between the zeroes and the coefficients.
Answer:
First, write the polynomial in standard form: \(p(x) = 6x^2 - 7x - 3\).
Factorize by splitting the middle term:
\(6x^2 - 9x + 2x - 3 = 0\)
\(3x(2x - 3) + 1(2x - 3) = 0\)
\((3x + 1)(2x - 3) = 0\)
So, the zeroes are \(x = -\frac{1}{3}\) and \(x = \frac{3}{2}\).
Verification:
Sum of zeroes = \(\frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6}\)
From coefficients, \(-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}\) (Verified)
Product of zeroes = \(\left(\frac{3}{2}\right)\left(-\frac{1}{3}\right) = -\frac{1}{2}\)
From coefficients, \(\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}\) (Verified)
In simple words: Factorize the quadratic equation to find roots, then check them against sum and product formulas.

Exam Tip: Always rearrange the polynomial in descending order of powers before attempting factorization.

Question 3. For what value of \(k\), \(-4\) is a zero of polynomial \(x^2 - x - (2k + 2)\)?
Answer:
Let \(p(x) = x^2 - x - (2k + 2)\).
Since \(-4\) is a zero of \(p(x)\), we must have \(p(-4) = 0\):
\((-4)^2 - (-4) - (2k + 2) = 0\)
\(16 + 4 - 2k - 2 = 0\)
\(20 - 2k - 2 = 0\)
\(18 - 2k = 0\)
\(2k = 18\)
\(k = 9\)
In simple words: Substitute \(x = -4\) into the polynomial and equate it to zero to solve for \(k\).

Exam Tip: Be very careful with parentheses and negative signs when substituting negative numbers.

Question 4. Give an example of polynomials \(p(x), g(x), q(x)\) and \(r(x)\) which satisfy division algorithm and \(\text{deg } p(x) = \text{deg } q(x)\).
Answer:
Let \(p(x) = 3x^2 + 2x + 1\), \(g(x) = 1\), \(q(x) = 3x^2 + 2x + 1\), and \(r(x) = 0\).
Check: \(p(x) = g(x) \cdot q(x) + r(x)\)
\(3x^2 + 2x + 1 = 1 \cdot (3x^2 + 2x + 1) + 0\) (Satisfied)
Here, \(\text{deg } p(x) = 2\) and \(\text{deg } q(x) = 2\), so \(\text{deg } p(x) = \text{deg } q(x)\).
In simple words: Choose a constant divisor so that the quotient has the exact same degree as the dividend.

Exam Tip: Degrees of polynomials are equal when the divisor is a non-zero constant polynomial.

Question 5. Find the zeroes of \(4u^2 + 8u\).
Answer:
Let \(p(u) = 4u^2 + 8u\).
Factor out the common terms: \(4u(u + 2) = 0\)
Setting each factor to zero:
\(4u = 0 \implies u = 0\)
\(u + 2 = 0 \implies u = -2\)
Thus, the zeroes are 0 and -2.
In simple words: Take out the greatest common factor instead of using middle-term splitting.

Exam Tip: For binomials lacking a constant term, simple factoring is faster than quadratic formulas.

Question 6. Find a quadratic polynomial, whose the sum and product of its zeroes are \(\frac{1}{4}, -1\).
Answer:
Given sum of zeroes \(\alpha + \beta = \frac{1}{4}\) and product of zeroes \(\alpha\beta = -1\).
The required quadratic polynomial is:
\(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
\(p(x) = x^2 - \frac{1}{4}x - 1\)
Multiplying by 4 to clear the fraction (optional standard form):
\(p(x) = 4x^2 - x - 4\)
In simple words: Substitute the given sum and product directly into the polynomial template.

Exam Tip: Multiplying by the denominator makes the polynomial look cleaner, though fractional polynomials are also valid.

( Level - 3 )

Question 1. Find the zeroes of polynomial \(x^3 - 2x^2 - x + 2\).
Answer:
Let \(p(x) = x^3 - 2x^2 - x + 2\).
By trial, test factors of constant term 2 (\(\pm 1, \pm 2\)):
\(p(1) = (1)^3 - 2(1)^2 - 1 + 2 = 1 - 2 - 1 + 2 = 0\)
So, \((x - 1)\) is a factor.
Divide \(p(x)\) by \(x - 1\) or group terms:
\(x^2(x - 2) - 1(x - 2) = 0\)
\((x^2 - 1)(x - 2) = 0\)
\((x - 1)(x + 1)(x - 2) = 0\)
Therefore, the zeroes are \(-1, 1, 2\).
In simple words: Find one zero by inspection, then factor the remaining quadratic expression.

Exam Tip: Grouping terms in pairs is often much faster than long division for cubic polynomials.

Question 2. If the zeroes of the polynomial \(x^3 - 3x^2 + x + 1\) are \(\alpha - \beta, \alpha, \alpha + \beta\). Find \(\alpha\) and \(\beta\).
Answer:
Let the zeroes be \(\alpha - \beta, \alpha, \alpha + \beta\).
Sum of zeroes = \((\alpha - \beta) + \alpha + (\alpha + \beta) = 3\alpha\)
From the polynomial \(x^3 - 3x^2 + x + 1\), sum of zeroes = \(-\frac{-3}{1} = 3\).
Therefore, \(3\alpha = 3 \implies \alpha = 1\).
Product of zeroes = \((\alpha - \beta)(\alpha)(\alpha + \beta) = \alpha(\alpha^2 - \beta^2)\)
From the polynomial, product of zeroes = \(-\frac{1}{1} = -1\).
Substitute \(\alpha = 1\):
\(1(1^2 - \beta^2) = -1\)
\(1 - \beta^2 = -1\)
\(\beta^2 = 2 \implies \beta = \pm\sqrt{2}\).
Thus, \(\alpha = 1\) and \(\beta = \pm\sqrt{2}\).
In simple words: Use the sum and product formulas for cubic polynomials to solve for the variables representing the roots.

Exam Tip: Assuming symmetric roots like \(\alpha-\beta, \alpha, \alpha+\beta\) simplifies algebraic manipulation significantly.

Question 3. Divide \(f(x) = 6x^3 + 11x^2 - 39x - 65\) by \(g(x) = x^2 - 1 + x\).
Answer:
Rearrange \(g(x) = x^2 + x - 1\).
Perform polynomial long division of \(6x^3 + 11x^2 - 39x - 65\) by \(x^2 + x - 1\):
1. Divide leading term \(6x^3\) by \(x^2\) to get \(6x\).
2. Multiply \(6x(x^2 + x - 1) = 6x^3 + 6x^2 - 6x\).
3. Subtract to get remainder \(5x^2 - 33x - 65\).
4. Divide leading term \(5x^2\) by \(x^2\) to get \(5\).
5. Multiply \(5(x^2 + x - 1) = 5x^2 + 5x - 5\).
6. Subtract to get final remainder \(-38x - 60\).
Quotient = \(6x + 5\), Remainder = \(-38x - 60\).
In simple words: Follow standard long division steps, matching degree by degree from highest to lowest.

Exam Tip: Always arrange both dividend and divisor in descending order of powers before dividing.

Question 4. Check whether the polynomial \(t^2 - 3\) is a factor of polynomial \(2t^4 + 3t^3 - 2t^2 - 9t - 12\) by applying the division algorithm.
Answer:
Divide \(2t^4 + 3t^3 - 2t^2 - 9t - 12\) by \(t^2 - 3\):
1. First term of quotient = \(2t^2\).
2. \(2t^2(t^2 - 3) = 2t^4 - 6t^2\). Subtract to get \(3t^3 + 4t^2 - 9t - 12\).
3. Next term of quotient = \(3t\).
4. \(3t(t^2 - 3) = 3t^3 - 9t\). Subtract to get \(4t^2 - 12\).
5. Next term of quotient = \(4\).
6. \(4(t^2 - 3) = 4t^2 - 12\). Subtract to get remainder 0.
Remainder = 0, Quotient = \(2t^2 + 3t + 4\).
Since the remainder is 0, the given polynomial is a factor.
In simple words: If the remainder after division is zero, the divisor is a factor of the dividend.

Exam Tip: State clearly at the end that the remainder being zero confirms the factor theorem.

( Level - 4 )

Question 1. Obtain all zeroes of \(f(x) = x^3 + 13x^2 + 32x + 20\).
Answer:
Let \(f(x) = x^3 + 13x^2 + 32x + 20\).
Test factors of 20: \(f(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 0\).
So, \((x + 1)\) is a factor.
Divide \(f(x)\) by \(x + 1\) using long division or synthetic division to get the quadratic factor:
\(x^3 + 13x^2 + 32x + 20 = (x + 1)(x^2 + 12x + 20)\)
Now factorize the quadratic part \(x^2 + 12x + 20\):
\(x^2 + 10x + 2x + 20 = x(x + 10) + 2(x + 10) = (x + 10)(x + 2)\)
Thus, all zeroes are \(-1, -2, -10\).
In simple words: Find one zero by trial, divide to get a quadratic equation, then factorize it completely.

Exam Tip: Always list all three zeroes clearly at the conclusion of a cubic polynomial problem.

Question 2. Obtain all other zeroes of \(3x^4 + 6x^3 - 2x^2 - 10x - 5\), if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\).
Answer:
Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes, the factors are:
\(\left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3}\)
Let \(g(x) = 3x^2 - 5\) (multiplying by 3 for convenience).
Divide the given polynomial by \(3x^2 - 5\):
1. Quotient term = \(x^2 + 2x + 1\).
2. Remainder = 0.
Now factorize the quotient polynomial \(x^2 + 2x + 1\):
\((x + 1)^2 = 0 \implies x = -1, -1\).
Thus, all other zeroes are \(-1\) and \(-1\).
In simple words: Construct a quadratic factor from the given roots, divide the main polynomial, and solve the resulting quotient.

Exam Tip: Repeated roots must be listed individually as stated by the multiplicity rules.

Question 3. On dividing \(x^3 - 3x^2 + x + 2\) by a polynomial \(g(x)\), the quotient and remainder were \(x - 2\) and \(-2x + 4\) respectively, find \(g(x)\).
Answer:
By division algorithm:
\(p(x) = g(x) \cdot q(x) + r(x)\)
\(x^3 - 3x^2 + x + 2 = g(x) \cdot (x - 2) + (-2x + 4)\)
Rearranging terms:
\(x^3 - 3x^2 + x + 2 - (-2x + 4) = g(x) \cdot (x - 2)\)
\(x^3 - 3x^2 + x + 2 + 2x - 4 = g(x) \cdot (x - 2)\)
\(x^3 - 3x^2 + 3x - 2 = g(x) \cdot (x - 2)\)
Now, divide \(x^3 - 3x^2 + 3x - 2\) by \(x - 2\):
1. Quotient = \(x^2 - x + 1\).
2. Remainder = 0.
Therefore, \(g(x) = x^2 - x + 1\).
In simple words: Rearrange the division algorithm formula to isolate \(g(x) \cdot q(x)\), then perform long division.

Exam Tip: Be very careful when distributing negative signs while transposing the remainder term.

(Problems for Self-Evaluation)

Question 1. Check whether \(g(x) = 3x - 2\) is a factor of \(p(x) = 3x^3 + x^2 - 20x + 12\).
Answer:
Perform long division of \(3x^3 + x^2 - 20x + 12\) by \(3x - 2\):
1. Quotient = \(x^2 + x - 6\).
2. Remainder = 0.
Since the remainder is 0, \(3x - 2\) is a factor of \(p(x)\).
In simple words: Divide the polynomial by the binomial; if the remainder is 0, it is a factor.

Exam Tip: Alternatively, use the factor theorem by substituting \(x = 2/3\) into \(p(x)\).

Question 2. Find quotient and remainder applying the division algorithm on dividing \(x^3 - 6x^2 + 2x - 4\) by \(g(x) = x - 1\).
Answer:
Divide \(x^3 - 6x^2 + 2x - 4\) by \(x - 1\):
1. Quotient = \(x^2 - 5x - 3\).
2. Remainder = -7.
In simple words: Divide term by term using polynomial long division until the degree of remainder is less than the divisor.

Exam Tip: Double check your subtraction signs at each step of the division process.

Question 3. Find zeros of the polynomial \(2x^2 - 8x + 6\).
Answer:
Let \(p(x) = 2x^2 - 8x + 6 = 2(x^2 - 4x + 3)\).
Factorize the quadratic expression inside:
\(x^2 - 3x - x + 3 = x(x - 3) - 1(x - 3) = (x - 1)(x - 3)\).
Setting \(p(x) = 0\):
\(2(x - 1)(x - 3) = 0\)
Thus, the zeroes are 1 and 3.
In simple words: Factor out the common coefficient first, then factorize the remaining quadratic.

Exam Tip: Factoring out common numerical factors simplifies the numbers you work with.

Question 4. Find the quadratic polynomial whose sum and product of its zeros are \(\frac{2}{3}, \frac{-1}{3}\) respectively.
Answer:
Given sum \(\alpha + \beta = \frac{2}{3}\) and product \(\alpha\beta = -\frac{1}{3}\).
Required polynomial:
\(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
\(p(x) = x^2 - \frac{2}{3}x - \frac{1}{3}\)
Multiplying by 3 for standard form:
\(p(x) = 3x^2 - 2x - 1\)
In simple words: Substitute the given sum and product into the quadratic template.

Exam Tip: Clearing denominators is standard practice when writing final polynomial answers.

Question 5. Find the zeroes of polynomial \(x^3 - 2x^2 - x + 2\).
Answer:
Let \(p(x) = x^3 - 2x^2 - x + 2\).
By grouping terms:
\(x^2(x - 2) - 1(x - 2) = 0\)
\((x^2 - 1)(x - 2) = 0\)
\((x - 1)(x + 1)(x - 2) = 0\)
Therefore, the zeroes are \(-1, 1, 2\).
In simple words: Group terms pairwise to factor the cubic expression directly.

Exam Tip: Always verify that all three factors multiply back to the original cubic polynomial.

Question 6. If one of the zeroes of the polynomial \(2x^2 + px + 4 = 0\) is 2, find the other root, also find the value of \(p\).
Answer:
Let \(\alpha = 2\) be one root of \(2x^2 + px + 4 = 0\).
Substitute \(x = 2\) to find \(p\):
\(2(2)^2 + p(2) + 4 = 0\)
\(8 + 2p + 4 = 0\)
\(12 + 2p = 0 \implies 2p = -12 \implies p = -6\).
Now, substitute \(p = -6\) into the equation: \(2x^2 - 6x + 4 = 0\).
Divide by 2: \(x^2 - 3x + 2 = 0\).
Factorize: \((x - 1)(x - 2) = 0\).
Thus, the other root is 1.
In simple words: Substitute the given root to find the unknown parameter, then solve the quadratic equation to find the second root.

Exam Tip: Alternatively, use the product of roots formula \(\alpha\beta = c/a\) to find the second root instantly.

Question 7. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(kx^2 + 4x + 4\), show that \(\alpha^2 + \beta^2 = 24\), find the value of \(k\).
Answer:
For \(kx^2 + 4x + 4\), sum of zeroes \(\alpha + \beta = -\frac{4}{k}\) and product of zeroes \(\alpha\beta = \frac{4}{k}\).
We know that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).
Substitute the values:
\(24 = \left(-\frac{4}{k}\right)^2 - 2\left(\frac{4}{k}\right)\)
\(24 = \frac{16}{k^2} - \frac{8}{k}\)
Multiply through by \(k^2\):
\(24k^2 = 16 - 8k\)
\(24k^2 + 8k - 16 = 0\)
Divide by 8: \(3k^2 + k - 2 = 0\)
Factorize: \(3k^2 + 3k - 2k - 2 = 0\)
\(3k(k + 1) - 2(k + 1) = 0\)
\((3k - 2)(k + 1) = 0\)
Thus, \(k = \frac{2}{3}\) or \(k = -1\).
In simple words: Use the algebraic identity for sum of squares and solve the resulting quadratic equation for \(k\).

Exam Tip: Memorize the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) as it appears frequently.

Question 8. If \(\alpha\) and \(\beta\) are the zeroes of the equation \(6x^2 + x - 2 = 0\), find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\).
Answer:
For \(6x^2 + x - 2 = 0\), comparing with \(ax^2 + bx + c\):
\(a = 6, \quad b = 1, \quad c = -2\)
Sum of zeroes: \(\alpha + \beta = -\frac{1}{6}\)
Product of zeroes: \(\alpha\beta = \frac{-2}{6} = -\frac{1}{3}\)
Now, simplify the target expression:
\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}\)
Substitute the values of sum and product:
\(= \frac{\left(-\frac{1}{6}\right)^2 - 2\left(-\frac{1}{3}\right)}{-\frac{1}{3}}\)
\(= \frac{\frac{1}{36} + \frac{2}{3}}{-\frac{1}{3}}\)
\(= \frac{\frac{1 + 24}{36}}{-\frac{1}{3}} = \frac{\frac{25}{36}}{-\frac{1}{3}}\)
\(= \frac{25}{36} \times \left(-\frac{3}{1}\right) = -\frac{25}{12}\)
In simple words: Convert the given fraction into an algebraic expression of sum and product of zeroes before substituting.

Exam Tip: Avoid finding individual roots \(\alpha\) and \(\beta\) when working with symmetric rational expressions of roots.