This MCQ module is based on: Relationship Between Zeroes and Coefficients
TOPIC 5 OF 35
Relationship Between Zeroes and Coefficients
🎓 Class 10
Mathematics
CBSE
Theory
Ch 2 — Polynomials
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Relationship Between Zeroes and Coefficients
Targeting Class 10 level in Algebra, with Intermediate difficulty.
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2.3 Relationship between Zeroes and Coefficients of a Polynomial
You have already seen that the zero of a linear polynomial \(ax + b\) is \(-\dfrac{b}{a}\). We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients? of a quadratic polynomial.
Deriving the Relationship for Quadratic Polynomials
Consider the quadratic polynomial \(p(x) = 2x^2 - 8x + 6\). To find its zeroes, we factorise by splitting the middle term. We need to split \(-8x\) as a sum of two terms whose product is \(2 \times 6 = 12x^2\):
\[2x^2 - 8x + 6 = 2x^2 - 6x - 2x + 6 = 2x(x - 3) - 2(x - 3) = 2(x - 1)(x - 3)\]So the value of \(p(x)\) is zero when \(x = 1\) or \(x = 3\). Therefore, the zeroes of \(2x^2 - 8x + 6\) are 1 and 3. Now observe:
\[\text{Sum of zeroes} = 1 + 3 = 4 = \frac{-(-8)}{2} = \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\] \[\text{Product of zeroes} = 1 \times 3 = 3 = \frac{6}{2} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]Let us take one more quadratic polynomial: \(p(x) = 3x^2 + 5x - 2\). By the method of splitting the middle term:
\[3x^2 + 5x - 2 = 3x^2 + 6x - x - 2 = 3x(x + 2) - 1(x + 2) = (3x - 1)(x + 2)\]So the zeroes are \(\frac{1}{3}\) and \(-2\). Now observe that:
\[\text{Sum of zeroes} = \frac{1}{3} + (-2) = \frac{1-6}{3} = \frac{-5}{3} = \frac{-(5)}{3} = \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\] \[\text{Product of zeroes} = \frac{1}{3} \times (-2) = \frac{-2}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\]General Result for Quadratic Polynomials
If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(p(x) = ax^2 + bx + c\), where \(a \ne 0\), then:
\[\alpha + \beta = \frac{-b}{a}\]
\[\alpha \cdot \beta = \frac{c}{a}\]
That is, sum of zeroes? \(= \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\), and product of zeroes? \(= \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\).
Derivation
Since \(\alpha\) and \(\beta\) are the zeroes of \(ax^2 + bx + c\), we have:
\[ax^2 + bx + c = a(x - \alpha)(x - \beta) = a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]\]Comparing the coefficients of \(x^2\), \(x\), and the constant terms on both sides, we get:
\[b = -a(\alpha + \beta) \quad \Rightarrow \quad \alpha + \beta = \frac{-b}{a}\] \[c = a \cdot \alpha\beta \quad \Rightarrow \quad \alpha\beta = \frac{c}{a}\]Example 2
Find the zeroes of the quadratic polynomial \(x^2 + 7x + 10\), and verify the relationship between the zeroes and the coefficients.
Solution: We have
\(x^2 + 7x + 10 = x^2 + 5x + 2x + 10 = x(x+5) + 2(x+5) = (x+2)(x+5)\)
So the value of \(x^2 + 7x + 10\) is zero when \(x + 2 = 0\) or \(x + 5 = 0\), i.e., when \(x = -2\) or \(x = -5\).
Therefore, the zeroes of \(x^2 + 7x + 10\) are \(-2\) and \(-5\). Now,
Sum of zeroes \(= -2 + (-5) = -7 = \dfrac{-(7)}{1} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= (-2) \times (-5) = 10 = \dfrac{10}{1} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
\(x^2 + 7x + 10 = x^2 + 5x + 2x + 10 = x(x+5) + 2(x+5) = (x+2)(x+5)\)
So the value of \(x^2 + 7x + 10\) is zero when \(x + 2 = 0\) or \(x + 5 = 0\), i.e., when \(x = -2\) or \(x = -5\).
Therefore, the zeroes of \(x^2 + 7x + 10\) are \(-2\) and \(-5\). Now,
Sum of zeroes \(= -2 + (-5) = -7 = \dfrac{-(7)}{1} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= (-2) \times (-5) = 10 = \dfrac{10}{1} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
Example 3
Find the zeroes of the polynomial \(x^2 - 3\), and verify the relationship between the zeroes and the coefficients.
Solution: Recall the identity \(a^2 - b^2 = (a-b)(a+b)\). Using it, we can write:
\(x^2 - 3 = \left(x - \sqrt{3}\right)\left(x + \sqrt{3}\right)\)
So the value of \(x^2 - 3\) is zero when \(x = \sqrt{3}\) or \(x = -\sqrt{3}\).
Therefore, the zeroes of \(x^2 - 3\) are \(\sqrt{3}\) and \(-\sqrt{3}\). Now,
Sum of zeroes \(= \sqrt{3} + (-\sqrt{3}) = 0 = \dfrac{-0}{1} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= \sqrt{3} \times (-\sqrt{3}) = -3 = \dfrac{-3}{1} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
\(x^2 - 3 = \left(x - \sqrt{3}\right)\left(x + \sqrt{3}\right)\)
So the value of \(x^2 - 3\) is zero when \(x = \sqrt{3}\) or \(x = -\sqrt{3}\).
Therefore, the zeroes of \(x^2 - 3\) are \(\sqrt{3}\) and \(-\sqrt{3}\). Now,
Sum of zeroes \(= \sqrt{3} + (-\sqrt{3}) = 0 = \dfrac{-0}{1} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= \sqrt{3} \times (-\sqrt{3}) = -3 = \dfrac{-3}{1} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
Example 4
Find the zeroes of the quadratic polynomial \(4s^2 - 4s + 1\), and verify the relationship between the zeroes and the coefficients.
Solution: We have
\(4s^2 - 4s + 1 = (2s - 1)^2\)
So the value of \(4s^2 - 4s + 1\) is zero when \(2s - 1 = 0\), i.e., \(s = \frac{1}{2}\).
Therefore, the zeroes of \(4s^2 - 4s + 1\) are \(\frac{1}{2}\) and \(\frac{1}{2}\) (a repeated zero). Now,
Sum of zeroes \(= \frac{1}{2} + \frac{1}{2} = 1 = \dfrac{-(-4)}{4} = \dfrac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^2}\) ✔
Product of zeroes \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \dfrac{1}{4} = \dfrac{\text{Constant term}}{\text{Coefficient of } s^2}\) ✔
\(4s^2 - 4s + 1 = (2s - 1)^2\)
So the value of \(4s^2 - 4s + 1\) is zero when \(2s - 1 = 0\), i.e., \(s = \frac{1}{2}\).
Therefore, the zeroes of \(4s^2 - 4s + 1\) are \(\frac{1}{2}\) and \(\frac{1}{2}\) (a repeated zero). Now,
Sum of zeroes \(= \frac{1}{2} + \frac{1}{2} = 1 = \dfrac{-(-4)}{4} = \dfrac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^2}\) ✔
Product of zeroes \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \dfrac{1}{4} = \dfrac{\text{Constant term}}{\text{Coefficient of } s^2}\) ✔
Example 5
Find the zeroes of the quadratic polynomial \(6x^2 - 3 - 7x\), and verify the relationship between the zeroes and the coefficients.
Solution: First, rearrange in standard form: \(6x^2 - 7x - 3\).
\(6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)\)
So the zeroes are \(x = -\frac{1}{3}\) and \(x = \frac{3}{2}\). Now,
Sum of zeroes \(= -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= \left(-\frac{1}{3}\right) \times \frac{3}{2} = -\frac{1}{2} = \dfrac{-3}{6} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
\(6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)\)
So the zeroes are \(x = -\frac{1}{3}\) and \(x = \frac{3}{2}\). Now,
Sum of zeroes \(= -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}\) ✔
Product of zeroes \(= \left(-\frac{1}{3}\right) \times \frac{3}{2} = -\frac{1}{2} = \dfrac{-3}{6} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}\) ✔
Example 6
Find a quadratic polynomial, the sum and product of whose zeroes are \(-3\) and \(2\), respectively.
Solution: Let the quadratic polynomial be \(ax^2 + bx + c\), and its zeroes be \(\alpha\) and \(\beta\).
We have: \(\alpha + \beta = -3 = \frac{-b}{a}\) and \(\alpha\beta = 2 = \frac{c}{a}\).
If \(a = 1\), then \(b = 3\) and \(c = 2\).
So one quadratic polynomial which fits the given conditions is \(x^2 + 3x + 2\).
Verification: \(x^2 + 3x + 2 = (x+1)(x+2)\), so zeroes are \(-1\) and \(-2\).
Sum = \(-1 + (-2) = -3\) ✔ Product = \((-1)(-2) = 2\) ✔
Note: Any polynomial of the form \(k(x^2 + 3x + 2)\), where \(k \ne 0\), is also valid. For example, \(2x^2 + 6x + 4\), \(-x^2 - 3x - 2\), etc.
We have: \(\alpha + \beta = -3 = \frac{-b}{a}\) and \(\alpha\beta = 2 = \frac{c}{a}\).
If \(a = 1\), then \(b = 3\) and \(c = 2\).
So one quadratic polynomial which fits the given conditions is \(x^2 + 3x + 2\).
Verification: \(x^2 + 3x + 2 = (x+1)(x+2)\), so zeroes are \(-1\) and \(-2\).
Sum = \(-1 + (-2) = -3\) ✔ Product = \((-1)(-2) = 2\) ✔
Note: Any polynomial of the form \(k(x^2 + 3x + 2)\), where \(k \ne 0\), is also valid. For example, \(2x^2 + 6x + 4\), \(-x^2 - 3x - 2\), etc.
Extension to Cubic Polynomials
Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients?
Let us consider \(p(x) = 2x^3 - 5x^2 - 14x + 8\). We can check that \(p(4) = 128 - 80 - 56 + 8 = 0\), \(p\!\left(-2\right) = -16 - 20 + 28 + 8 = 0\), and \(p\!\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{5}{4} - 7 + 8 = 0\). So the three zeroes are \(4\), \(-2\), and \(\frac{1}{2}\). Now observe:
\[\text{Sum of zeroes} = 4 + (-2) + \frac{1}{2} = \frac{5}{2} = \frac{-(-5)}{2} = \frac{-(\text{Coefficient of } x^2)}{\text{Coefficient of } x^3}\] \[\text{Sum of products taken two at a time} = 4(-2) + (-2)\!\left(\frac{1}{2}\right) + 4\!\left(\frac{1}{2}\right) = -8 - 1 + 2 = -7 = \frac{-14}{2} = \frac{\text{Coefficient of } x}{\text{Coefficient of } x^3}\] \[\text{Product of zeroes} = 4 \times (-2) \times \frac{1}{2} = -4 = \frac{-(8)}{2} = \frac{-(\text{Constant term})}{\text{Coefficient of } x^3}\]General Result for Cubic Polynomials
If \(\alpha\), \(\beta\), \(\gamma\) are the zeroes of the cubic polynomial \(ax^3 + bx^2 + cx + d\), where \(a \ne 0\), then:
\[\alpha + \beta + \gamma = \frac{-b}{a}\]
\[\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\]
\[\alpha\beta\gamma = \frac{-d}{a}\]
Example 7
Verify that \(3\), \(-1\), and \(-\frac{1}{3}\) are the zeroes of the cubic polynomial \(p(x) = 3x^3 - 5x^2 - 11x - 3\), and then verify the relationship between the zeroes and the coefficients.
Solution: Comparing with \(ax^3 + bx^2 + cx + d\), we get \(a = 3\), \(b = -5\), \(c = -11\), \(d = -3\).
Verifying zeroes:
\(p(3) = 3(27) - 5(9) - 11(3) - 3 = 81 - 45 - 33 - 3 = 0\) ✔
\(p(-1) = 3(-1) - 5(1) - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0\) ✔
\(p\!\left(-\frac{1}{3}\right) = 3\!\left(-\frac{1}{27}\right) - 5\!\left(\frac{1}{9}\right) - 11\!\left(-\frac{1}{3}\right) - 3 = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = -\frac{6}{9} + \frac{11}{3} - 3 = -\frac{2}{3} + \frac{11}{3} - 3 = \frac{9}{3} - 3 = 0\) ✔
Therefore, \(3\), \(-1\), and \(-\frac{1}{3}\) are indeed zeroes of \(p(x)\).
So we take \(\alpha = 3\), \(\beta = -1\), \(\gamma = -\frac{1}{3}\). Now,
\(\alpha + \beta + \gamma = 3 + (-1) + \left(-\frac{1}{3}\right) = 2 - \frac{1}{3} = \frac{5}{3} = \frac{-(-5)}{3} = \frac{-b}{a}\) ✔
\(\alpha\beta + \beta\gamma + \gamma\alpha = 3(-1) + (-1)\!\left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right)(3) = -3 + \frac{1}{3} - 1 = \frac{-9+1-3}{3} = \frac{-11}{3} = \frac{c}{a}\) ✔
\(\alpha\beta\gamma = 3 \times (-1) \times \left(-\frac{1}{3}\right) = 1 = \frac{-(-3)}{3} = \frac{-d}{a}\) ✔
Verifying zeroes:
\(p(3) = 3(27) - 5(9) - 11(3) - 3 = 81 - 45 - 33 - 3 = 0\) ✔
\(p(-1) = 3(-1) - 5(1) - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0\) ✔
\(p\!\left(-\frac{1}{3}\right) = 3\!\left(-\frac{1}{27}\right) - 5\!\left(\frac{1}{9}\right) - 11\!\left(-\frac{1}{3}\right) - 3 = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = -\frac{6}{9} + \frac{11}{3} - 3 = -\frac{2}{3} + \frac{11}{3} - 3 = \frac{9}{3} - 3 = 0\) ✔
Therefore, \(3\), \(-1\), and \(-\frac{1}{3}\) are indeed zeroes of \(p(x)\).
So we take \(\alpha = 3\), \(\beta = -1\), \(\gamma = -\frac{1}{3}\). Now,
\(\alpha + \beta + \gamma = 3 + (-1) + \left(-\frac{1}{3}\right) = 2 - \frac{1}{3} = \frac{5}{3} = \frac{-(-5)}{3} = \frac{-b}{a}\) ✔
\(\alpha\beta + \beta\gamma + \gamma\alpha = 3(-1) + (-1)\!\left(-\frac{1}{3}\right) + \left(-\frac{1}{3}\right)(3) = -3 + \frac{1}{3} - 1 = \frac{-9+1-3}{3} = \frac{-11}{3} = \frac{c}{a}\) ✔
\(\alpha\beta\gamma = 3 \times (-1) \times \left(-\frac{1}{3}\right) = 1 = \frac{-(-3)}{3} = \frac{-d}{a}\) ✔
Exercise 2.2
Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) \(x^2 - 2x - 8\)
\(x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x-4) + 2(x-4) = (x+2)(x-4)\)
Zeroes: \(x = -2\) and \(x = 4\).
Sum of zeroes \(= -2 + 4 = 2 = \frac{-(-2)}{1} = \frac{-b}{a}\) ✔
Product of zeroes \(= (-2)(4) = -8 = \frac{-8}{1} = \frac{c}{a}\) ✔
Zeroes: \(x = -2\) and \(x = 4\).
Sum of zeroes \(= -2 + 4 = 2 = \frac{-(-2)}{1} = \frac{-b}{a}\) ✔
Product of zeroes \(= (-2)(4) = -8 = \frac{-8}{1} = \frac{c}{a}\) ✔
(ii) \(4s^2 - 4s + 1\)
\(4s^2 - 4s + 1 = (2s - 1)^2\)
Zeroes: \(s = \frac{1}{2}\) and \(s = \frac{1}{2}\) (repeated zero).
Sum of zeroes \(= \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-b}{a}\) ✔
Product of zeroes \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{c}{a}\) ✔
Zeroes: \(s = \frac{1}{2}\) and \(s = \frac{1}{2}\) (repeated zero).
Sum of zeroes \(= \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-b}{a}\) ✔
Product of zeroes \(= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{c}{a}\) ✔
(iii) \(6x^2 - 3 - 7x\)
Rearranging: \(6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)\)
Zeroes: \(x = -\frac{1}{3}\) and \(x = \frac{3}{2}\).
Sum of zeroes \(= -\frac{1}{3} + \frac{3}{2} = \frac{-2+9}{6} = \frac{7}{6} = \frac{-(-7)}{6} = \frac{-b}{a}\) ✔
Product of zeroes \(= \left(-\frac{1}{3}\right)\!\left(\frac{3}{2}\right) = -\frac{1}{2} = \frac{-3}{6} = \frac{c}{a}\) ✔
Zeroes: \(x = -\frac{1}{3}\) and \(x = \frac{3}{2}\).
Sum of zeroes \(= -\frac{1}{3} + \frac{3}{2} = \frac{-2+9}{6} = \frac{7}{6} = \frac{-(-7)}{6} = \frac{-b}{a}\) ✔
Product of zeroes \(= \left(-\frac{1}{3}\right)\!\left(\frac{3}{2}\right) = -\frac{1}{2} = \frac{-3}{6} = \frac{c}{a}\) ✔
(iv) \(4u^2 + 8u\)
\(4u^2 + 8u = 4u(u + 2)\)
Zeroes: \(u = 0\) and \(u = -2\).
Sum of zeroes \(= 0 + (-2) = -2 = \frac{-(8)}{4} = \frac{-b}{a}\) ✔
Product of zeroes \(= 0 \times (-2) = 0 = \frac{0}{4} = \frac{c}{a}\) ✔
Zeroes: \(u = 0\) and \(u = -2\).
Sum of zeroes \(= 0 + (-2) = -2 = \frac{-(8)}{4} = \frac{-b}{a}\) ✔
Product of zeroes \(= 0 \times (-2) = 0 = \frac{0}{4} = \frac{c}{a}\) ✔
(v) \(t^2 - 15\)
\(t^2 - 15 = \left(t - \sqrt{15}\right)\left(t + \sqrt{15}\right)\)
Zeroes: \(t = \sqrt{15}\) and \(t = -\sqrt{15}\).
Sum of zeroes \(= \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a}\) ✔
Product of zeroes \(= \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a}\) ✔
Zeroes: \(t = \sqrt{15}\) and \(t = -\sqrt{15}\).
Sum of zeroes \(= \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a}\) ✔
Product of zeroes \(= \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a}\) ✔
(vi) \(3x^2 - x - 4\)
\(3x^2 - x - 4 = 3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4)\)
Zeroes: \(x = -1\) and \(x = \frac{4}{3}\).
Sum of zeroes \(= -1 + \frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3} = \frac{-(-1)}{3} = \frac{-b}{a}\) ✔
Product of zeroes \(= (-1) \times \frac{4}{3} = -\frac{4}{3} = \frac{-4}{3} = \frac{c}{a}\) ✔
Zeroes: \(x = -1\) and \(x = \frac{4}{3}\).
Sum of zeroes \(= -1 + \frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3} = \frac{-(-1)}{3} = \frac{-b}{a}\) ✔
Product of zeroes \(= (-1) \times \frac{4}{3} = -\frac{4}{3} = \frac{-4}{3} = \frac{c}{a}\) ✔
Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), \(-1\)
Sum \(= \frac{1}{4}\), Product \(= -1\).
The quadratic polynomial is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - \frac{1}{4}x - 1\).
Multiplying by 4: \(\boxed{4x^2 - x - 4}\)
The quadratic polynomial is \(x^2 - (\text{sum})x + (\text{product}) = x^2 - \frac{1}{4}x - 1\).
Multiplying by 4: \(\boxed{4x^2 - x - 4}\)
(ii) \(\sqrt{2}\), \(\frac{1}{3}\)
Sum \(= \sqrt{2}\), Product \(= \frac{1}{3}\).
The quadratic polynomial is \(x^2 - \sqrt{2}\,x + \frac{1}{3}\).
Multiplying by 3: \(\boxed{3x^2 - 3\sqrt{2}\,x + 1}\)
The quadratic polynomial is \(x^2 - \sqrt{2}\,x + \frac{1}{3}\).
Multiplying by 3: \(\boxed{3x^2 - 3\sqrt{2}\,x + 1}\)
(iii) \(0\), \(\sqrt{5}\)
Sum \(= 0\), Product \(= \sqrt{5}\).
The quadratic polynomial is \(x^2 - 0 \cdot x + \sqrt{5} = \boxed{x^2 + \sqrt{5}}\)
The quadratic polynomial is \(x^2 - 0 \cdot x + \sqrt{5} = \boxed{x^2 + \sqrt{5}}\)
(iv) \(1\), \(1\)
Sum \(= 1\), Product \(= 1\).
The quadratic polynomial is \(\boxed{x^2 - x + 1}\)
The quadratic polynomial is \(\boxed{x^2 - x + 1}\)
(v) \(-\frac{1}{4}\), \(\frac{1}{4}\)
Sum \(= -\frac{1}{4}\), Product \(= \frac{1}{4}\).
The quadratic polynomial is \(x^2 + \frac{1}{4}x + \frac{1}{4}\).
Multiplying by 4: \(\boxed{4x^2 + x + 1}\)
The quadratic polynomial is \(x^2 + \frac{1}{4}x + \frac{1}{4}\).
Multiplying by 4: \(\boxed{4x^2 + x + 1}\)
(vi) \(4\), \(1\)
Sum \(= 4\), Product \(= 1\).
The quadratic polynomial is \(\boxed{x^2 - 4x + 1}\)
The quadratic polynomial is \(\boxed{x^2 - 4x + 1}\)
2.4 Summary
Chapter Summary
- Polynomials of degrees 1, 2, and 3 are called linear, quadratic, and cubic polynomials respectively.
- A quadratic polynomial in \(x\) with real coefficients is of the form \(ax^2 + bx + c\), where \(a\), \(b\), \(c\) are real numbers with \(a \ne 0\).
- The zeroes of a polynomial \(p(x)\) are precisely the x-coordinates of the points where the graph of \(y = p(x)\) intersects the x-axis.
- A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.
- If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(ax^2 + bx + c\), then:
\(\alpha + \beta = \dfrac{-b}{a}\), \(\alpha\beta = \dfrac{c}{a}\). - If \(\alpha\), \(\beta\), \(\gamma\) are the zeroes of the cubic polynomial \(ax^3 + bx^2 + cx + d\), then:
\(\alpha + \beta + \gamma = \dfrac{-b}{a}\), \(\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a}\), \(\alpha\beta\gamma = \dfrac{-d}{a}\).
Activity: Discover the Sum-Product Pattern
L4 Analyse
Materials: Pencil, paper, calculator (optional)
Predict: If you know the zeroes of a quadratic are 5 and -2, can you write the polynomial without any formula? Think about how \((x - 5)(x + 2)\) expands.
- Choose any two integers as zeroes of a quadratic, say \(\alpha\) and \(\beta\).
- Write the quadratic as \((x - \alpha)(x - \beta)\) and expand it to get \(x^2 - (\alpha+\beta)x + \alpha\beta\).
- Now compute \(\frac{-b}{a}\) and \(\frac{c}{a}\) from the expanded form. Do they match the sum and product?
- Repeat with three different pairs of zeroes. What pattern do you observe?
Observation: For any pair of zeroes \(\alpha, \beta\), when we write \((x-\alpha)(x-\beta) = x^2 - (\alpha+\beta)x + \alpha\beta\), we get \(a=1\), \(b = -(\alpha+\beta)\), \(c = \alpha\beta\). Then \(\frac{-b}{a} = \alpha+\beta\) and \(\frac{c}{a} = \alpha\beta\). The relationships always hold because factorisation and expansion are inverse operations!
Interactive: Zeroes-Coefficients Verifier
Enter a quadratic \(ax^2 + bx + c\). The tool finds the zeroes and verifies the sum/product relationships.
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Competency-Based Questions
Scenario: A rectangular garden has a length that is 3 metres more than twice its width. If the area of the garden is 27 m², the dimensions can be found using the quadratic polynomial \(2w^2 + 3w - 27\), where \(w\) is the width in metres.
Q1. What are the zeroes of \(2w^2 + 3w - 27\)?
L3 ApplyAnswer: (a). \(2w^2 + 3w - 27 = 2w^2 + 9w - 6w - 27 = w(2w+9) - 3(2w+9) = (w-3)(2w+9)\). So \(w = 3\) or \(w = -\frac{9}{2}\).
Q2. Verify that the sum and product of the zeroes satisfy the expected relationships with the coefficients.
L3 ApplyAnswer: Sum \(= 3 + \left(-\frac{9}{2}\right) = -\frac{3}{2} = \frac{-(3)}{2} = \frac{-b}{a}\) ✔
Product \(= 3 \times \left(-\frac{9}{2}\right) = -\frac{27}{2} = \frac{-27}{2} = \frac{c}{a}\) ✔
Product \(= 3 \times \left(-\frac{9}{2}\right) = -\frac{27}{2} = \frac{-27}{2} = \frac{c}{a}\) ✔
Q3. Why must we reject the negative zero in this context? Explain what it would mean geometrically for the garden problem.
L4 AnalyseAnswer: Width must be a positive real number. \(w = -\frac{9}{2}\) is rejected because a negative width has no physical meaning. The valid width is \(w = 3\) m, giving length \(= 2(3)+3 = 9\) m. Geometrically, \(w = -\frac{9}{2}\) is the other x-intercept of the parabola \(y = 2w^2+3w-27\), which lies in the negative half of the number line — outside the domain of valid widths.
Q4. Create a new word problem that leads to a quadratic polynomial with zeroes whose sum is 7 and product is 12. Write the polynomial and state the problem context.
L6 CreateSample Answer: Polynomial: \(x^2 - 7x + 12 = (x-3)(x-4)\), zeroes are 3 and 4.
Problem: "Two consecutive integers differ by 1. Their sum is 7 and their product is 12. Find the integers." Or: "A rectangular plot has dimensions (in metres) that are the zeroes of \(x^2 - 7x + 12\). Find the perimeter." (Perimeter = 2(3+4) = 14 m.)
Problem: "Two consecutive integers differ by 1. Their sum is 7 and their product is 12. Find the integers." Or: "A rectangular plot has dimensions (in metres) that are the zeroes of \(x^2 - 7x + 12\). Find the perimeter." (Perimeter = 2(3+4) = 14 m.)
Assertion-Reason Questions
Assertion (A): If the zeroes of a quadratic polynomial are both positive, then the coefficients \(b\) and \(c\) (in \(ax^2 + bx + c\), \(a > 0\)) have the same sign.
Reason (R): Sum of zeroes \(= -b/a\) and product of zeroes \(= c/a\).
Reason (R): Sum of zeroes \(= -b/a\) and product of zeroes \(= c/a\).
Answer: (d) — A is false. If both zeroes are positive, then their sum is positive, so \(-b/a > 0\), meaning \(b < 0\) (when \(a > 0\)). Their product is positive, so \(c/a > 0\), meaning \(c > 0\). So \(b\) is negative and \(c\) is positive — they have opposite signs. R is true.
Assertion (A): The quadratic polynomial \(x^2 - 5x + 6\) has zeroes 2 and 3.
Reason (R): For \(ax^2 + bx + c\) with zeroes \(\alpha, \beta\): the polynomial can be written as \(a(x - \alpha)(x - \beta)\).
Reason (R): For \(ax^2 + bx + c\) with zeroes \(\alpha, \beta\): the polynomial can be written as \(a(x - \alpha)(x - \beta)\).
Answer: (a) — \(x^2-5x+6 = (x-2)(x-3)\), confirming zeroes are 2 and 3. R is true and explains A: we can factor \(p(x)\) as \((x-2)(x-3)\), directly showing the zeroes.
Frequently Asked Questions — Polynomials
What is Relationship Between Zeroes and Coefficients in NCERT Class 10 Mathematics?
Relationship Between Zeroes and Coefficients is a key concept covered in NCERT Class 10 Mathematics, Chapter 2: Polynomials. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Relationship Between Zeroes and Coefficients step by step?
To solve problems on Relationship Between Zeroes and Coefficients, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Polynomials?
The essential formulas of Chapter 2 (Polynomials) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Relationship Between Zeroes and Coefficients important for the Class 10 board exam?
Relationship Between Zeroes and Coefficients is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Relationship Between Zeroes and Coefficients?
Common mistakes in Relationship Between Zeroes and Coefficients include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Relationship Between Zeroes and Coefficients?
End-of-chapter NCERT exercises for Relationship Between Zeroes and Coefficients cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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