Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
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\(x^2 – 2x – 8\\\)
Solution:
\(\\ \text{Given:} \quad x^2 - 2x - 8 \\\\ \text{Find two numbers whose product is }\\ a \times c = 1 \times (-8) = -8, \\ \\\text{and whose sum is } b = -2 \\\\ \text{The suitable pair is } \\\\2 \text{ and } -4\ \text{since}\ 2 \times (-4) \\\\= -8\ \text{and}\ 2 + (-4) = -2. \\\\ \\\) Split the middle term and factor:
\begin{aligned}&x^2 - 2x - 8 = x^2 + 2x - 4x - 8 \\ &= (x^2 + 2x) + (-4x - 8) \\ &= x(x + 2) - 4(x + 2) \\ &= (x + 2)(x - 4) \\ \\ &\text{Step 3: Find zeros:} \\ &x + 2 = 0 \implies x = -2 \\ &x - 4 = 0 \implies x = 4 \\ \\ &\text{Verification:} \\ &\text{Sum of roots: } 4 + (-2) = 2 \\ &-\frac{b}{a} = -\frac{-2}{1} = 2 \\ \\ &\text{Product of roots: } 4 \times (-2) = -8 \\ &\frac{c}{a} = \frac{-8}{1} = -8 \\ \end{aligned} Hence, the factorisation and roots are correct. -
\(4s^2 – 4s + 1\\\)
Solution:
\(\textbf{Factorization of } 4s^2 - 4s + 1\) \[4s^2 - 4s + 1 \] We rewrite the middle term: \[4s^2 - 2s - 2s + 1\] Now, we factor by grouping: \[2s(2s - 1) - 1(2s - 1)\] \[= (2s - 1)(2s - 1)\] Setting each factor equal to zero, we find the roots: \[2s - 1 = 0 \implies s = \frac{1}{2}\] Thus, both roots are \[s = \frac{1}{2}\] \(\textbf{Verification: \\\\Sum of roots}\) \[\frac{1}{2} + \frac{1}{2} = \frac{-b}{a}\] \[1 = \frac{-(-4)}{4}\] \[ 1 = 1\] The sum of the roots is correct.
\(\textbf{Verification: \\\\Product of roots}\) \[\frac{1}{2} \times \frac{1}{2} = \frac{c}{a}\] \[\frac{1}{4} = \frac{1}{4}\] The product of the roots is correct. - \(
6x^2 – 3 – 7x\\\)
Solution:
\(\text{\textbf{Given quadratic:}} \quad p(x) = 6x^2 - 7x - 3 \\[1em]\\ \text{Here,} \quad a = 6,\quad b = -7,\quad c = -3 \\[1em]\\ \text{Find two numbers whose product is } \\\\a \times c = 6 \times (-3) = -18,\\\\ \text{and whose sum is } b = -7. \\\\[0.7em] \text{These numbers are } -9 \text{ and } +2.\\\\\) \begin{align} (-9) + (2) &= -7 \\\\ (-9) \times (2) &= -18\\\end{align} \(\text{Split the middle term:} \) \begin{align*}p(x) &= 6x^2 - 9x + 2x - 3 \\\end{align*} \(\text{Group and factor:} \\\) \begin{align*}&= (6x^2 - 9x) + (2x - 3) \\ &= 3x(2x - 3) + 1(2x - 3) \\ &= (2x - 3)(3x + 1)\\[1em]\end{align*} \(\text{Set each factor equal to zero:} \\\) \begin{align*}2x - 3 = 0 &\implies x = \frac{3}{2} \\ 3x + 1 = 0 &\implies x = -\frac{1}{3} \\[1em]\end{align*} \(\text{\textbf{Roots: }} \left(\frac{3}{2},\ -\frac{1}{3}\right)\\[1em]\\\\\) \(\text{\textbf{Verification:}} \\\\[0.8em] \text{Sum of roots:} \\\) \begin{align*}\frac{3}{2} + \left(-\frac{1}{3}\right) &= \frac{9 - 2}{6} = \frac{7}{6} \\ -\frac{b}{a} &= -\frac{-7}{6} = \frac{7}{6} \\\end{align*} \(\textit{Sum is verified.} \\\\[1em] \text{Product of roots: }\\\) \begin{align*}\frac{3}{2} \times \left(-\frac{1}{3}\right) &= -\frac{3}{6} = -\frac{1}{2} \\ \frac{c}{a} &= \frac{-3}{6} = -\frac{1}{2} \\\end{align*} \(\text{Product is verified.}\) -
\(4u2 + 8u\\\)
\[ \textbf{Given:} \quad p(u) = 4u^2 + 8u \] Factorise the quadratic expression: \[ \begin{align*} p(u) &= 4u^2 + 8u \\ &= 4u(u + 2) \end{align*} \] Find the roots Let \( p(u) = 0 \): \[ \begin{align*} 4u(u+2) &= 0 \\ \implies u = 0 \quad\text{or}\quad u + 2 = 0 \\ \implies u = 0 \quad\text{or}\quad u = -2 \end{align*} \] Roots: \(\quad 0 \quad\text{and } -2\\\\\) Verification: \[ \text{Sum of roots: Let } \alpha = 0,\, \beta = -2 \] \[ \begin{align*} \alpha + \beta &= 0 + (-2) = -2 \\\\ -\frac{b}{a} &= -\frac{8}{4} = -2 \\\\ \text{Sum verified:} &\quad -2 = -2 \end{align*} \] \( \text{Product of roots:} \) \[ \begin{align*} \alpha \beta &= 0 \times (-2) = 0 \\\\ \frac{c}{a} &= \frac{0}{4} = 0 \\\\ 0 &= 0 \end{align*} \] \(\text{Product verified:}\)
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\(t^2 – 15\\\)
\[ \textbf{Given:}\quad p(t) = t^2 - 15 \] \(\text{Find the roots}\) \[ \begin{align*} p(t) = 0 &\implies t^2 - 15 = 0 \\ &\implies t^2 = 15 \\ &\implies t = \pm \sqrt{15} \end{align*} \] \(\textbf{Roots:}\quad \sqrt{15},\ -\sqrt{15}\\\\ \text{Verification:\\} \text{Sum of roots:}\) \[ \begin{align*} \sqrt{15} + (-\sqrt{15}) &= 0 \\\\ -\frac{b}{a} &= -\frac{0}{1} = 0 \\\\ \text{Sum verified:}\quad 0 = 0 \end{align*} \] \(\text{Product of roots:}\) \[ \begin{align*} \sqrt{15} \times (-\sqrt{15}) &= -15 \\\\ \frac{c}{a} &= \frac{-15}{1} \\&= -15 \\\\ \text{Product verified:}\quad -15 &= -15 \end{align*} \]
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\( 3x^2 – x – 4\\\)
\[ \textbf{Given:}\quad p(x) = 3x^2 - x - 4 \] Split the Middle Term \[ a \times c = 3 \times (-4) = -12 \] \[ \text{Factors of}\ -12\ \text{with sum}\ -1:\\\\ -4\ \text{and}\ +3 \] \[ p(x) = 3x^2 - 4x + 3x - 4 \] Factor by Grouping:
\( 3x^2 - 4x + 3x - 4 \\= (3x^2 - 4x) + (3x - 4) \\\) \(\begin{aligned}&= x(3x - 4) + 1(3x - 4) \\ &= (3x - 4)(x + 1) \end{aligned} \)\[\] Find the Roots: \[\begin{aligned} 3x - 4 &= 0 \\\implies x &= \frac{4}{3} \\\\ x + 1 &= 0 \\\implies x &= -1 \end{aligned}\] \[ \text{Roots:}\quad \frac{4}{3},\; -1 \] Verify the Roots:
Sum of Roots \[ \begin{align*} \alpha+\beta&=\frac{4}{3} + (-1) \\\\&= \frac{4-3}{3} \\\\&= \frac{1}{3} \\\\ -\frac{b}{a} &= -\frac{-1}{3} \\\\&= \frac{1}{3} \end{align*} \] \[ \text{Sum of roots is verified.} \] Product of Roots \[ \begin{align*} \alpha\beta&=\frac{c}{a}\\\\ \frac{4}{3} \times (-1) &= -\frac{4}{3} \\\\ \frac{c}{a} &= \frac{-4}{3} \end{align*} \] \( \text{Product of roots is verified.} \)
Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
- \(\frac{1}{4}, -1\\\\\)Solution:
\[ \textbf{Let}\quad p(x) = ax^2 + bx + c \] \(\text{Suppose}\quad \alpha,\ \beta \text{ are roots of }p(x)\) \[\begin{aligned} \alpha+\beta &= -\frac{b}{a} \\&= \frac{1}{4}\\\\ \implies -\frac{b}{a} &= \frac{1}{4}\\\\ \implies a &= -4b\\\\ \text{Also}\quad \alpha\beta &= \frac{c}{a} \\\\&= -1\\ \implies \frac{c}{a} &= -1 \\\implies c &= -a \end{aligned}\] \[ \text{From above:}\quad a = -4b,\quad c = -a \] \[\begin{aligned} \text{Let}\ b &= -1,\\ a &= -4 \times (-1) = 4,\\ c &= -4 \end{aligned}\] \[ \textbf{Quadratic Equation:} \] \[ \begin{align*} p(x) &= ax^2 + bx + c \\ &= 4x^2 - x - 4 \end{align*}\] \[ \text{Sum of roots:} \] \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\ &= -\frac{-1}{4} = \frac{1}{4} \end{align*} \] \[ \text{Verified:}\quad \frac{1}{4} = \frac{1}{4} \] \[ \text{Product of roots:} \] \[ \begin{align*} \alpha\beta &= \frac{c}{a} \\ &= \frac{-4}{4} = -1 \end{align*} \] \[ \text{Verified:}\quad -1 = -1 \] - \(\sqrt(2), \frac{1}{3}\\\\\)Solution:
\[ \textbf{Let the quadratic be}\\p(x) = ax^2 + bx + c \] Given:
Sum of roots: \[\begin{aligned} \alpha + \beta &= -\frac{b}{a}\\ \alpha + \beta &= \sqrt{2}\\ \implies \sqrt{2} &= -\frac{b}{a} \\\implies a &= -\frac{b}{\sqrt{2}} \end{aligned} \] Product of roots: \[\begin{aligned} \alpha \beta &= \frac{c}{a}\\\\ \alpha \beta &= \frac{1}{3}\\\\ \implies \frac{1}{3} &= \frac{c}{a} \\\\ \implies a &= 3c \end{aligned}\] Solving for: \(a,\ b,\ c:\\\\\) Assume: \[\begin{aligned} a &= 1\\b &= -\sqrt{2},\\ c &= \frac{1}{3} \end{aligned}\] So, the quadratic equation: \[ \begin{align*} p(x) &= ax^2 + bx + c \\ &= x^2 + (-\sqrt{2})x + \frac{1}{3} \\ &= x^2 - \sqrt{2}x + \frac{1}{3} \end{align*} \] \(\textbf{Verification of quadratic equation: } p(x)\\\\ \text{Sum of roots:}\) \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\ &= -\frac{-\sqrt{2}}{1} \\ &= \sqrt{2} \end{align*} \] Verified:\[\sqrt{2} = \sqrt{2}\] Product of roots: \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\ &= \frac{1/3}{1} \\ &= \frac{1}{3} \end{align*} \] Verified: \[ \frac{1}{3} = \frac{1}{3} \] - \(0,\sqrt{5}\\\\\)Solution:
Let Quadratic Equation be \(p(x)=ax^2+bx+c\\\\\) Sum of roots, say \(\left(\alpha, \beta \right)\\\) \[ \begin{aligned} \alpha+\beta\ &=-\frac{b}{a}\\\\ 0&=-\frac{b}{a}\\\\ a\cdot 0&=-b\\ \implies b&=0, \quad a \ne0 \end{aligned} \] Product of roots \[\begin{aligned} \alpha\beta&=\frac{c}{a}\\\\ \sqrt{5}&=\frac{c}{a}\\\\ a&=\frac{c}{\sqrt{5}} \end{aligned}\] Assume: \[\begin{aligned} a&=1\\b&=0\\c&=\sqrt{5} \end{aligned}\] \[\begin{aligned} p(x)&=x^2+0\cdot x + \sqrt{5}\\ &=x^2+\sqrt{5} \end{aligned}\] Verification of Sum of Roots: \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\ &= -\frac{0}{1} \\ &= 0 \end{align*} \] Verified:\[\begin{aligned} 0 &= 0 \end{aligned}\] Product of Roots: \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\ &= \frac{\sqrt{5}}{1} \\ &= \sqrt{5} \end{align*} \] \[ \text{Verified:} \quad \sqrt{5} = \sqrt{5} \] - \(1,1\\\\\)Solution:
\( \textbf{Let the quadratic equation be} \\\\p(x) = ax^2 + bx + c\\\\ \text{Given:} \\\\ \alpha + \beta = 1, \\\alpha\beta = 1\\\\ \text{Use the sum of roots formula} \) \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\\\ 1 &= -\frac{b}{a} \\\\ \implies a &= -b \end{align*} \] \( \text{Use the product of roots formula} \) \[ \begin{align*} \alpha\beta &= \frac{c}{a} \\\\ 1 &= \frac{c}{a} \\\\ \implies a &= c \end{align*} \] \( \text{Substitute} \) \[ a = -b, \;\; a = c \] If \[\begin{aligned} a &= 1\\ b &= -1, \\ c &= 1 \end{aligned}\] \( \textbf{The quadratic equation is} \) \[ p(x) = x^2 - x + 1 \] \( \text{Verify the sum of roots} \) \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\\\ &= -\frac{-1}{1} \\\\ &= 1 \end{align*} \] \( \text{Verified:} \; 1 = 1 \) \[ \text{Verify the product of roots} \] \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\\\ &= \frac{1}{1} \\\\ &= 1 \end{align*} \] \( \text{Verified:} \; 1 = 1\\\\ \) Hence, for \(\quad p(x) = x^2 - x + 1\\\\\) both sum and product of roots are correctly verified. - \(-\frac{1}{4},\frac{1}{4}\\\\\)Solution:
\( \textbf{Let the quadratic equation be:} \\\\ p(x) = ax^2 + bx + c \) Given: \[\begin{aligned} \alpha + \beta &= -\frac{b}{a} \\\\&= -\frac{1}{4} \\\\ \alpha\beta &= \frac{c}{a} \\\\&= \frac{1}{4}\end{aligned} \] Use sum and product formula \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\ -\frac{1}{4} &= -\frac{b}{a} \\ a &= 4b \end{align*} \] \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\ \frac{1}{4} &= \frac{c}{a} \\ a &= 4c \end{align*} \] \[ \text{So,} \quad a = 4b = 4c \] Assign values \[\begin{aligned}\text{if } a &= 1, \text{ then }\\ b &= \frac{1}{4} \\ c &= \frac{1}{4} \end{aligned}\] Form the quadratic polynomial \[ p(x) = x^2 + \frac{1}{4}x + \frac{1}{4} \] Verify sum of roots \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\ &= -\frac{1/4}{1} \\ &= -\frac{1}{4} \end{align*} \] \( \text{Verified:} \quad -\frac{1}{4} = -\frac{1}{4} \) Verify the product of roots \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\\\ &= \frac{\left(1/4\right)}{1} \\\\ &= \frac{1}{4} \end{align*} \] Verified: \(\quad \dfrac{1}{4} = \dfrac{1}{4}\\\\\) for \[ p(x) = x^2 + \frac{1}{4}x + \frac{1}{4} \] - \(4,1\\\\\)Solution:
\( \textbf{Let the quadratic polynomial be} \\ p(x) = ax^2 + bx + c \\\) \(\begin{aligned} \text{Given:} \\\alpha + \beta &= 4 \\\alpha \beta &= 1 \end{aligned}\\\) \( \text{Sum of roots} \) \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\\\ 4 &= -\frac{b}{a} \\\\ 4a &= -b \\ b &= -4a \end{align*} \] \( \text{Product of roots formula} \) \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\\\ 1 &= \frac{c}{a} \\\\ c &= a \end{align*} \] \( \text{Combine results}\\\\ b = -4a,\; c = a \\\\\) Let \[\begin{aligned}a &= 1 \text{ then}\\ b &= -4,\\ c &= 1 \end{aligned}\] \( \text{Form the polynomial} \) \[ p(x) = x^2 - 4x + 1 \] Verification \( \\\\\text{Sum of roots:} \) \[ \begin{align*} \alpha + \beta &= -\frac{b}{a} \\\\ &= -\frac{-4}{1} \\\\ &= 4 \end{align*} \] \( \text{Product of roots:} \) \[ \begin{align*} \alpha \beta &= \frac{c}{a} \\\\ &= \frac{1}{1} \\\\ &= 1 \end{align*} \] \[ \boxed{p(x) = x^2 - 4x + 1} \]
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