EXERCISE 2.2
Polynomials • 2 Questions
Question 1
Hint available
Find the zeroes of the following quadratic and verify the relationship between the zeroes and the coefficients. (i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) 4u2 + 8u (v) t2 – 15 (vi) 3x2 – x – 4
Key Idea
For a quadratic polynomial $ax^2+bx+c=0$, if its zeroes are $\alpha$ and $\beta$, then \[\alpha+\beta = -\frac{b}{a} \quad\text{and}\quad \alpha\beta = \frac{c}{a}.\] This relationship must be verified after finding the zeroes.
Step-by-Step Solution
1. General method – For each quadratic we either factorise (if possible) or use the quadratic formula \[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.\] After obtaining the two zeroes $\alpha$ and $\beta$, compute \(\alpha+\beta\) and \(\alpha\beta\) and compare with \(-b/a\) and \(c/a\) respectively.\
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(i) $x^2-2x-8$\
- $a=1,\;b=-2,\;c=-8$.\
- Factorising: $x^2-2x-8=(x-4)(x+2)$.\
- Zeroes: $\alpha=4$, $\beta=-2$.\
- Sum $\alpha+\beta = 4+(-2)=2 = -\frac{b}{a}= -\frac{-2}{1}=2$.\
- Product $\alpha\beta = 4\times(-2) = -8 = \frac{c}{a}=\frac{-8}{1}=-8$.\
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(ii) $4s^2-4s+1$\
- $a=4,\;b=-4,\;c=1$.\
- Discriminant $\Delta = (-4)^2-4\cdot4\cdot1 = 0$.\
- Double root: $s = \frac{-b}{2a}=\frac{4}{8}=\frac12$.\
- Zeroes: $\alpha=\beta=\frac12$.\
- Sum $\alpha+\beta = 1 = -\frac{b}{a}= -\frac{-4}{4}=1$.\
- Product $\alpha\beta = \frac14 = \frac{c}{a}=\frac{1}{4}$.\
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(iii) $6x^2-7x-3$ (re‑ordered from $6x^2-3-7x$)\
- $a=6,\;b=-7,\;c=-3$.\
- $\Delta = (-7)^2-4\cdot6\cdot(-3)=49+72=121$.\
- Roots: $x = \frac{7\pm\sqrt{121}}{12}=\frac{7\pm11}{12}$.\
\[\alpha = \frac{18}{12}=\frac32,\qquad \beta = \frac{-4}{12}= -\frac13.\]\
- Sum $\alpha+\beta = \frac32-\frac13 = \frac{7}{6}= -\frac{b}{a}= -\frac{-7}{6}=\frac{7}{6}$.\
- Product $\alpha\beta = \frac32\times\left(-\frac13\right)= -\frac12 = \frac{c}{a}=\frac{-3}{6}= -\frac12$.\
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(iv) $4u^2+8u$\
- $a=4,\;b=8,\;c=0$.\
- Factor: $4u(u+2)=0$.\
- Zeroes: $\alpha=0$, $\beta=-2$.\
- Sum $\alpha+\beta = -2 = -\frac{b}{a}= -\frac{8}{4}= -2$.\
- Product $\alpha\beta = 0 = \frac{c}{a}=\frac{0}{4}=0$.\
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(v) $t^2-15$\
- $a=1,\;b=0,\;c=-15$.\
- Roots: $t = \pm\sqrt{15}$.\
- Zeroes: $\alpha=\sqrt{15}$, $\beta=-\sqrt{15}$.\
- Sum $\alpha+\beta = 0 = -\frac{b}{a}=0$.\
- Product $\alpha\beta = -15 = \frac{c}{a}= -15$.\
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(vi) $3x^2 - x - 4$\
- $a=3,\;b=-1,\;c=-4$.\
- $\Delta = (-1)^2-4\cdot3\cdot(-4)=1+48=49$.\
- Roots: $x = \frac{1\pm7}{6}$.\
\[\alpha = \frac{8}{6}=\frac43,\qquad \beta = \frac{-6}{6}= -1.\]\
- Sum $\alpha+\beta = \frac43-1 = \frac13 = -\frac{b}{a}= -\frac{-1}{3}=\frac13$.\
- Product $\alpha\beta = \frac43\times(-1)= -\frac43 = \frac{c}{a}=\frac{-4}{3}= -\frac43$.
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(i) $x^2-2x-8$\
- $a=1,\;b=-2,\;c=-8$.\
- Factorising: $x^2-2x-8=(x-4)(x+2)$.\
- Zeroes: $\alpha=4$, $\beta=-2$.\
- Sum $\alpha+\beta = 4+(-2)=2 = -\frac{b}{a}= -\frac{-2}{1}=2$.\
- Product $\alpha\beta = 4\times(-2) = -8 = \frac{c}{a}=\frac{-8}{1}=-8$.\
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(ii) $4s^2-4s+1$\
- $a=4,\;b=-4,\;c=1$.\
- Discriminant $\Delta = (-4)^2-4\cdot4\cdot1 = 0$.\
- Double root: $s = \frac{-b}{2a}=\frac{4}{8}=\frac12$.\
- Zeroes: $\alpha=\beta=\frac12$.\
- Sum $\alpha+\beta = 1 = -\frac{b}{a}= -\frac{-4}{4}=1$.\
- Product $\alpha\beta = \frac14 = \frac{c}{a}=\frac{1}{4}$.\
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(iii) $6x^2-7x-3$ (re‑ordered from $6x^2-3-7x$)\
- $a=6,\;b=-7,\;c=-3$.\
- $\Delta = (-7)^2-4\cdot6\cdot(-3)=49+72=121$.\
- Roots: $x = \frac{7\pm\sqrt{121}}{12}=\frac{7\pm11}{12}$.\
\[\alpha = \frac{18}{12}=\frac32,\qquad \beta = \frac{-4}{12}= -\frac13.\]\
- Sum $\alpha+\beta = \frac32-\frac13 = \frac{7}{6}= -\frac{b}{a}= -\frac{-7}{6}=\frac{7}{6}$.\
- Product $\alpha\beta = \frac32\times\left(-\frac13\right)= -\frac12 = \frac{c}{a}=\frac{-3}{6}= -\frac12$.\
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(iv) $4u^2+8u$\
- $a=4,\;b=8,\;c=0$.\
- Factor: $4u(u+2)=0$.\
- Zeroes: $\alpha=0$, $\beta=-2$.\
- Sum $\alpha+\beta = -2 = -\frac{b}{a}= -\frac{8}{4}= -2$.\
- Product $\alpha\beta = 0 = \frac{c}{a}=\frac{0}{4}=0$.\
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(v) $t^2-15$\
- $a=1,\;b=0,\;c=-15$.\
- Roots: $t = \pm\sqrt{15}$.\
- Zeroes: $\alpha=\sqrt{15}$, $\beta=-\sqrt{15}$.\
- Sum $\alpha+\beta = 0 = -\frac{b}{a}=0$.\
- Product $\alpha\beta = -15 = \frac{c}{a}= -15$.\
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(vi) $3x^2 - x - 4$\
- $a=3,\;b=-1,\;c=-4$.\
- $\Delta = (-1)^2-4\cdot3\cdot(-4)=1+48=49$.\
- Roots: $x = \frac{1\pm7}{6}$.\
\[\alpha = \frac{8}{6}=\frac43,\qquad \beta = \frac{-6}{6}= -1.\]\
- Sum $\alpha+\beta = \frac43-1 = \frac13 = -\frac{b}{a}= -\frac{-1}{3}=\frac13$.\
- Product $\alpha\beta = \frac43\times(-1)= -\frac43 = \frac{c}{a}=\frac{-4}{3}= -\frac43$.
Question 2
Hint available
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. (i) 1 , 1 4 (ii) 1 2 , 3 (iii) 0, 5 (iv) 1, 1 (v) 1 1 , 4 4 (vi) 4, 1
Key Idea
For a quadratic polynomial $ax^2+bx+c=0$ with zeroes $\alpha$ and $\beta$, the sum and product of the zeroes are given by \[\alpha+\beta = -\frac{b}{a},\qquad \alpha\beta = \frac{c}{a}.\] Choosing $a=1$ (monic polynomial) gives the convenient form \[x^2-(\text{sum})x+\text{product}=0.\] Multiplying by a suitable integer removes fractions, if required.
Step-by-Step Solution
1. Write the general monic quadratic: \[x^2 - Sx + P = 0,\] where $S$ is the required sum of the zeroes and $P$ is the required product.\
2. Substitute the given values of $S$ and $P$ for each part and simplify.\
3. If the coefficients are fractional, multiply the whole equation by the least common multiple of the denominators to obtain an equivalent polynomial with integer coefficients (the polynomial is still valid because multiplying by a non‑zero constant does not change its zeroes).\
4. State the final polynomial (either in the monic form or the cleared‑denominator form).\
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(i) Sum $S=1$, Product $P=-\frac{1}{4}$\
\[x^2 - 1x - \frac{1}{4}=0\]\
Multiply by $4$: \[4x^2 - 4x - 1 = 0.\]\
(ii) Sum $S=\frac{1}{2}$, Product $P=3$\
\[x^2 - \frac{1}{2}x + 3 = 0\]\
Multiply by $2$: \[2x^2 - x + 6 = 0.\]\
(iii) Sum $S=0$, Product $P=5$\
\[x^2 - 0x + 5 = 0 \;\Rightarrow\; x^2 + 5 = 0.\]\
(iv) Sum $S=1$, Product $P=1$\
\[x^2 - x + 1 = 0.\]\
(v) Sum $S=1$, Product $P=-1$ (since $4\,4^{-}\;= -1$)\
\[x^2 - x - 1 = 0.\]\
(vi) Sum $S=4$, Product $P=1$\
\[x^2 - 4x + 1 = 0.\]
2. Substitute the given values of $S$ and $P$ for each part and simplify.\
3. If the coefficients are fractional, multiply the whole equation by the least common multiple of the denominators to obtain an equivalent polynomial with integer coefficients (the polynomial is still valid because multiplying by a non‑zero constant does not change its zeroes).\
4. State the final polynomial (either in the monic form or the cleared‑denominator form).\
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(i) Sum $S=1$, Product $P=-\frac{1}{4}$\
\[x^2 - 1x - \frac{1}{4}=0\]\
Multiply by $4$: \[4x^2 - 4x - 1 = 0.\]\
(ii) Sum $S=\frac{1}{2}$, Product $P=3$\
\[x^2 - \frac{1}{2}x + 3 = 0\]\
Multiply by $2$: \[2x^2 - x + 6 = 0.\]\
(iii) Sum $S=0$, Product $P=5$\
\[x^2 - 0x + 5 = 0 \;\Rightarrow\; x^2 + 5 = 0.\]\
(iv) Sum $S=1$, Product $P=1$\
\[x^2 - x + 1 = 0.\]\
(v) Sum $S=1$, Product $P=-1$ (since $4\,4^{-}\;= -1$)\
\[x^2 - x - 1 = 0.\]\
(vi) Sum $S=4$, Product $P=1$\
\[x^2 - 4x + 1 = 0.\]