Question

 If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then

a = -7, b = -1

a = 5, b = -1

a = 2, b = -6

a =0, b = -6

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Answer 1

Answer:

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Answer 2

Answer:

(iv) a =0, b = -6

Step-by-step explanation:

$\large{\underline{\underline{\mathfrak{Given:-}}}}</p><p>$

zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3.

$\large{\underline{\underline{\mathfrak{To\:Find:-}}}}</p><p>$

value of a and b

$\large{\underline{\underline{\mathfrak{Concepts:-}}}}</p><p>$

Roots of the equations should satisfy the equation

Roots , zeroes and value are same thing

$\large{\underline{\underline{\mathfrak{Solution:-}}}}</p><p>$

So, we are given 2 roots 2, -3, we make 2 equation and find the value of a and b

Put the value of x as 2

2²+(a+1)2+b=0

4+2a+2+b=0

6+2a+b=0 -----------(1)

Put the value of x as -3

(-3)²+(a+1)(-3)+b=0

9-3a-3+b=0

6-3a+b=0 -----------(2)

Equate equation 1 and 2,

6+2a+b=6-3a+b=0

2a+3a=0

5a=0

a=0

We found an important information

Put the value of x and a ( any value of x =2 or -3 )

2²+(a+1)2+b=0

4+2a+2+b=0

4+2+b=0

6+b=0

b= -6

$\large{\underline{\underline{\mathfrak{Useful\: Formulae:-}}}}</p><p>$

• Sum of the zeroes= -b/a
• Product of the zeroes= c/a
• general form of a quadratic equation= ax²+bx+c=0