Question

Advait’s mother gave him some money to buy Papaya from the market at the rate of p(x) = x2 – 12x – 220. Let α, β are the zeroes of p(x). Based on the above information, answer the following questions. Q1. Find the values of α and β, where α < β. a) –22 and 10 b) –10 and 22 c) 2 and 14 d) –20 and 8 Q2. Find the value of α + β + αβ. a) 232 b) -232 c) 208 d) –208 Q3. The value of p(4) is a) –268 b) –168 c) –252 d) –164​

Answer 1

Instead of alpha and beta , I am taking it as a and b.
x2-12x-220=0
x2-22x+10x-220=0
x(x-22)+10(x-22)=0
(x-22)(x+10)=0
x=22 or x=-10

Q1) Value of alpha and beta is 22 and -10

Q2)a+b+ab=22+-10+22X-10=22-10-220=
-208

Q3)4X4-12X4-220=16-48-220= -252

Answer 2

Given is a quadratic equation $p(x)=x^2-12x-220$. Find its zeroes and other related values.

Explanation:

• Let there be a quadratic equation of the form $ax^2+bx+c$ having its zeroes $\alpha,\ \beta$ such that $\alpha<\beta$.
• Then using the quadratic formula we have the zeros as, $\alpha=\frac{-b-\sqrt{b^2-4ac} }{2a}\ \ \ \ \beta= \frac{-b+\sqrt{b^2-4ac} }{2a}$                           -----------(a) $->\alpha+\beta+\alpha\beta=\frac{c-b}{a}$                                            -----------(b)
• Now here we have $p(x)=x^2-12x-220$
• Hence we have, $a=1,\ b=-12,\ c=-220$  
• From (a) and (b) we get,
• ANSWER.1:
•                 $\alpha=\frac{12-\sqrt{144+880} }{2}\ \ \ \ \beta= \frac{12+\sqrt{144+880} }{2}\\\\->\alpha=-10,\ \beta=22$ ->Option(b)
• ANSWER.2:
•               $\alpha+\beta+\alpha\beta=\frac{-220+12}{1}\\->\alpha+\beta+\alpha\beta=-208$           ->Option(d)
• ANSWER.3:
•               $p(x)=x^2-12x-220\\p(4)=4^2-12(4)-220\\->p(4)=-252$            ->Option(c)