(i) $x^2 – 2x – 8$
(ii) $4s^2 – 4s + 1$
(iii) $6x^2 – 3 – 7x$
(iv) $4u^2+8u$
(v) $t^2$ – 15
(vi) $3x^2 – x – 4$
Relationship between the zeroes and the coefficients
| Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$ = $\dfrac{-(Coefficient\:\:of\:\:x)} {(Coefficient \:\:of \:\:x^{2})}$ |
| Product of zeroes = $\alpha \times \beta $ =$\dfrac{c}{a}$ = $\dfrac{Constant \:term} {(Coefficient \:\: of \:\: x^{2})}$ |
(i) $x^2 – 2x – 8$
For p(x) = $x^2 – 2x – 8$ = 0
$x^2– 4x+2x–8$ = 0
x(x–4)+2(x–4) = 0
(x-4)(x+2)=0
(x-4)=0 => x = 4
(x+2)=0 => x = -2
Therefore, zeroes of polynomial equation $x^2–2x–8$ are (4, -2)
$\alpha$ = 4 , $\beta$ = -2
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
4-2 = $\dfrac{-2}{1}$
2 = 2
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$4\times(-2)$ = $\dfrac{-8}{1}$
-8 = -8
Relationship between the zeroes and the coefficients are verified
(ii) $4s^2 – 4s + 1$
For p(s) = $4s^2 – 4s + 1$ = 0
$4s^2–2s–2s+1$ = 0
2s(2s–1)–1(2s-1) = 0
(2s–1)(2s–1)=0
(2s−1) = 0 => s = $\dfrac{1}{2}$
(2s−1) = 0 => s = $\dfrac{1}{2}$
Therefore, zeroes of polynomial equation $4s^2–4s+1$ are $(\dfrac{1}{2}, \dfrac{1}{2})$
$\alpha$ = $\dfrac{1}{2}$, $\beta$ = $\dfrac{1}{2}$
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
$(\dfrac{1}{2})+(\dfrac{1}{2}) = -\frac{-4}{4}$
1=1
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$(\dfrac{1}{2})\times(\dfrac{1}{2}) = \dfrac{1}{4}$
$\dfrac{1}{4}$ = $\dfrac{1}{4}$
Relationship between the zeroes and the coefficients are verified
(iii) $6x^2 – 3 – 7x$
For p(x) = $6x^2 – 3 – 7x$ = 0
$6x^2–7x–3$ = 0
$6x^2 – 9x + 2x – 3$ = 0
3x(2x – 3) +1(2x – 3) =0
(3x+1)(2x-3) = 0
(3x+1) = 0 => x = $\dfrac{-1}{3}$
(2x-3) = 0 => x = $\dfrac{3}{2}$
Therefore, zeroes of polynomial equation $6x^2 – 3 – 7x$ are $\dfrac{-1}{3} , \dfrac{3}{2}$
$\alpha$ = $\dfrac{-1}{3}$, $\beta$ = $\dfrac{3}{2}$
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
$\dfrac{-1}{3} + \dfrac{3}{2}$ = $\dfrac{-(-7)}{6}$
$\dfrac{7}{6}$ = $\dfrac{7}{6}$
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$\dfrac{-1}{3} \times \dfrac{3}{2}$ = $\dfrac{-3}{6}$
$\dfrac{-3}{6}$ = $\dfrac{-3}{6}$
Relationship between the zeroes and the coefficients are verified
(iv) $4u^2+8u$
For p(u) = $4u^2+8u$ = 0
$4u^2 + 8u$ = 0
4u(u+2) = 0
u = 0, u= -2
Therefore, zeroes of polynomial equation $4u^2 + 8u$ are 0,-2
$\alpha$ = 0, $\beta$ = -2
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
0 + (-2) = $\dfrac{-8}{4}$
(-2) = (-2)
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$0 \times (-2)$ = $\dfrac{0}{4}$
0 = 0
Relationship between the zeroes and the coefficients are verified
(v) $t^2$ – 15
For p(t) = $t^2$ – 15 = 0
$t^2$ – 15 = 0
$t^2$ = 15
t = $\pm \sqrt{15}$
$\alpha$ = + $\sqrt{15}$,$\beta$ = - $\sqrt{15}$
Therefore, zeroes of polynomial equation $t^2$ – 15 are $\pm \sqrt{15}$
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
$\sqrt{15}$ + (- $\sqrt{15}$) = $\dfrac{-0}{1}$
0 = 0
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$\sqrt{15} \times (-\sqrt{15})$ = $\dfrac{-15}{1}$
-15 = -15
Relationship between the zeroes and the coefficients are verified
(vi) $3x^2 – x – 4$
For p(x) = $3x^2 – x – 4$ = 0
$3x^2$–4x+3x–4 = 0
x(3x-4)+1(3x-4) = 0
(3x – 4)(x + 1) = 0
(3x – 4) = 0 => x = $\dfrac{4}{3}$
(x + 1) = 0 => x = -1
zeroes of the polynomial equation $3x^2 – x – 4$ are ($\dfrac{4}{3}$, -1)
$\alpha$ = $\dfrac{4}{3}$, $\beta$ = -1
Relationship between the zeroes and the coefficients
Sum of zeroes = $\alpha + \beta = \dfrac{-b}{a}$
$\dfrac{4}{3}$ + (-1) = $\dfrac{-(-1)}{3}$
$\dfrac{1}{3}$ = $\dfrac{1}{3}$
Product of zeros = $\alpha \times \beta $ =$\dfrac{c}{a}$
$\dfrac{4}{3} \times$ (-1) = $\dfrac{-4}{3}$
$\dfrac{-4}{3}$ = $\dfrac{-4}{3}$
Relationship between the zeroes and the coefficients are verified
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i)$\dfrac{1}{4}$,-1(ii) $\sqrt{2}, \dfrac{1}{3}$
(iii) 0, $\sqrt{5}$
(iv) 1, 1
(v) $\dfrac{-1}{4},\dfrac{1}{4}$
(vi) 4, 1
General form of Quadratic Polynomial
| $x^2$ − (Sum of zeroes)x + (Product of zeroes) |
(i)$\dfrac{1}{4}$,-1
Sum of zeroes = α+β = $\dfrac{1}{4}$
Product of zeroes = αβ = -1
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
= $x^2$–$\dfrac{1}{4}$x +(-1)
= $4x^2–\dfrac{4x}{4}-{4}$
= $4x^2$–x-4
Thus, the quadratic polynomial is 4x^2–x–4.
(ii) $\sqrt{2}, \dfrac{1}{3}$
Sum of zeroes = α+β = $\sqrt{2}$
Product of zeroes = αβ = $\dfrac{1}{3}$
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
=$x^2–\sqrt{2}x +\dfrac{1}{3}$
=$3x^2–3 \sqrt{2}x +\dfrac{3}{3}$
=${3x^2–3 \sqrt{2}x +1}$
Thus, the quadratic polynomial is ${3x^2–3 \sqrt{2}x +1}$
(iii) 0, $\sqrt{5}$
Sum of zeroes = α+β = 0
Product of zeroes = αβ = $\sqrt{5}$
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
=$x^2$–(0)x + $\sqrt{5}$
=$x^2$ + $\sqrt{5}$
Thus, the quadratic polynomial is $x^2$ + $\sqrt{5}$
(iv) 1, 1
Sum of zeroes = α+β = 1
Product of zeroes = αβ = 1
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
=$x^2$– 1 x +1
=$x^2$– x + 1
Thus, the quadratic polynomial is $x^2$– x + 1
(v) $\dfrac{-1}{4},\dfrac{1}{4}$
Sum of zeroes = α+β = $\dfrac{-1}{4}$
Product of zeroes = αβ = $\dfrac{1}{4}$
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
=$x^2 – \dfrac{-1}{4}x +\dfrac{1}{4}$
=$4x^2+\dfrac{4x}{4}+\dfrac{4}{4}$
=$4x^2$+x+1
Thus, the quadratic polynomial is $4x^2$+x+1
(vi) 4, 1
Sum of zeroes = α+β = 4
Product of zeroes = αβ = 1
Using the General form of Quadratic Polynomial
= $x^2$–(α+β)x +αβ
=$x^2$–4x + 1
Thus, the quadratic polynomial is $x^2$–4x + 1