Question

(a)a rainbow is represented by the quadratic polynomial x^2+(a+1)x+b whose zeroes are 2 and -3 . then
(I) a=7, b= +1 (ii) a=5 ,b=-1 (iii) a =2, b = -6 (iv) a=0, b= -6​

(b) the polynomial x^2-2x-(7p+3) represents a rainbow. if, then the value of p is
(I) 1 (ii) 2 (iii) 3 (iv) 4
plzz tell me fast

Answer 1

Answer:

2+(-3) =-(a+1) /1

-1=-a-1

a=0

2(-3) =b/1

-6=b

b=-6

Answer 2

A quadratic polynomial is a polynomial of second degree with the highest degree term equal to 2. $ax^{2} + bx + c = 0$ is the general form of a quadratic equation. The coefficients are a and b, the unknown variable is x, and the constant term is c.

Given:

$x^{2} +(a+1)x+b$ whose zeroes are $2$ and $-3$.

(a) A polynomial's zeroes are all the $x$ values at which the polynomial equals zero.

$p(x) = x^{2} + (a + 1)x + b$ i.e. $p(2) = 0$and $p(- 3) = 0$.

$p(2) = 2^{2} (2)2 + (a + 1)(2) + b = 0 \\\Rightarrow 4 + 2a + 2 + b = 0 \\\Rightarrow 6 + 2a + b = 0 ......(i) \\P(- 3) = (-3)^{2} + 9 + (a + 1)(- 3) + b = 0 \\\Rightarrow 9 - 3a - 3 + b = 0 \\\Rightarrow 6 - 3a + b = 0 .....(ii)$

Since both equations are equal to zero. As a result, the two equations are equal. 

$6 + 2a + b = 6 - 3a + b \\\Rightarrow 5a = 0\\\Rightarrow a = 0$

Substituting the value of ‘$a$’ in (i)

$6 + 2(0) + b = 0 \\\Rightarrow b = - 6$

Correct Answer is (iv)$a=0,b=-6$.

(b) Given:

Polynomial $x^2-2x-(7p+3)$

If $-4$ is zero of given polynomial then

$(-4)^{2} -2(-2)-(7p+3)=0\\\Rightarrow 16+4-7p=3=0\\\Rightarrow 7p=21\\\Rightarrow p=3$

Correct Answer is (iii)