Question

- 32. Without finding the zeroes a and ß of the polynomial p(x) = 6x2 - 13x + 6, find the values of
$\alpha ^{2} + \beta {}^{2}$

$1 \div \alpha + 1 \div \beta$

Answer 1

Answer:

97/36, 1/6

Step-by-step explanation:

here, sum of roots or alpha + beta = -b/a = 13/6

and peoduct of roots or alpha x beta = c/a = 6/6 = 1

so alpha^2 + beta ^2 = (alpha + beta)^2 - 2alpha x beta

= (13/6)^2 - 2x1 = 97/36

also, 1/alpha + 1/beta = sum of roots/product

= 13/6 ÷ 1 = 13/6

Answer 2

$Given, the polynomial </p><p></p><p>6x2−13x+6=0⇒6x2−9x−4x+6=0⇒3x(2x−3)−2(3x−3)=0⇒(2x−3)(3x−2)=0∴2x−3=0⇒x= \frac{3}{2} 3x−2=0 ⇒x= \frac{2}{3} </p><p></p><p>Let, α= \frac{3}{2} ,β= \frac{2}{3} </p><p></p><p>From the polynomial we get, a=6,b=−13,c=6</p><p></p><p>∴ Relation between the zeroes and coefficients,</p><p></p><p>Sum of the zeroes= \frac{3}{2} + \frac{2}{3} = \frac{13}{6} = - \frac{ - 13}{6} = \frac{b}{a} </p><p></p><p>and product of the zeroes=2 \frac{3}{2} × \frac{2}{3} = \frac{6}{6} = \frac{c}{a} </p><p></p><p>Hence, the relation between the zeroes and the coefficients of the polynomial is verified.</p><p></p><p></p><p>$