Question

16. If -2 and 3 are the zeros of the quadratic polynomial x2 + (a +1)x+b​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Correct Question:-

If -2 and 3 are the zeros of the quadratic polynomial $\tt x^2 + (a +1)x+b$ then find the value of a and b.

Answer:-

The Required Value Of a and b is -2 and -6.

Explanation:-

Given:-

• Quadratic Polynomial $\tt x^2+(a+1)x + b.$

• -2 and 3 are zeroes of the polynomial.

To Find:-

• The Value of a and b.

Concept Used:-

In a quadratic polynomial $\tt p(x) = ax^2+bx+c.$ if $\tt\alpha\:and\:\beta$ are zeroes of the polynomial then,

$\\ \tt\mapsto \alpha + \beta = - \frac{b}{a} .$

$\\ \tt\mapsto \alpha \beta = \frac{c}{a} .$

Solution:-

So Here,

$\\ \tt\longrightarrow \alpha = - 2.$

$\\ \tt\longrightarrow \beta = 3$

$\\ \tt\longrightarrow a = 1.$

$\\ \tt\longrightarrow b = (a + 1)$

$\\ \tt\longrightarrow c = b.$

Therefore,

$\\ \tt\mapsto \alpha + \beta = - \frac{b}{a} .$

$\\ \tt\mapsto( - 2) + 3 = - \frac{(a + 1)}{1} .$

$\\ \tt\mapsto - a - 1 = 1.$

$\\ \tt\mapsto - a = 2.$

$\\ \red{\large\mapsto \boxed{ \blue{ \bf a = - 2.}}}$

Also,

$\\ \tt\mapsto \alpha \beta = \frac{c}{a} .$

$\\ \tt\mapsto ( - 2) \times (3) = \frac{b}{1} .$

$\\ \red{\large\mapsto\boxed{ \blue{ \bf b = - 6.}}}$

Therefore The Required Value of a and b is -2 and -6 respectively.