Question

10. If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then *
O(a) a = -7, b = -1
O (b) a = 5, b = -1
0 (c) a = 2, b = -6
(d) a = 0, b = -6​

Answer 1

Given:

• The Zeroes of the Quadratic Polynomial x² + (a + 1) x + b are 2 and – 3.
• Options are given, we've to choose the correct option.

To find:

• The Value of a and b?

Solution: It is given that, 2 and – 3 are the Zeroes of the Quadratic polynomial x² + (a + 1)x + b. So, we'll substitute the value of x as 2 and – 3.

• Putting x = 2.

➟ x² + (a + 1)x + b = 0

➟ (2)² + (a + 1)2 + b = 0

➟ 4 + 2a + 2 + b = 0

➟ 6 + 2a + b = 0 ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀( eqⁿ (i) )

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

• Putting x = – 3.

➟ x² + (a + 1)x + b = 0

➟ (–3)² + (a + 1)(–3) + b = 0

➟ 9 – 3a – 3 + b = 0

➟ 6 – 3a + b = 0⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ( eqⁿ (ii) )

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

• Now, comparing eqⁿ ( i ) and eqⁿ ( ii ) —

➟ 6 + 2a + b = 6 – 3a + b

➟ 5a = 0

➟ a = 0

• Substituting the value of a in eqⁿ ( i ) —

➟ 6 + 2a + b = 0

➟ 6 + 2(0) + b = 0

➟ 6 + b = 0

➟ b = – 6

$\therefore$ Hence$,$ the values of a and b are Option d) 0 and – 6.

Answer 2

Answer:

Question :-

• If the zeroes of the quadratic polynomial
• ${x}^{2} + (a + 1)x + ab$
• Are 2 and -3.Then
• (i)a=-7 and b=-1
• (ii)a=5 and b=-1
• (iii)a=2 and b=-6
• (iv)a=0 and b=-6

$\mathcal\pink{♧Given:-}$

• In the question given that the 2 and -3 are the zeroes of the Quadratic polynomial
• ${x}^{2} + (a + 1)x + b$

$\mathcal\purple{♧To prove:-}$

• Here we should find the value of a and b.

$\mathcal\pink{♧Explanation: -}$

• Here first we substitute the value of x as 2 and -3 in given quadratic polynomial.

$\mathcal\pink{♧Now,}$

• Taking x as 2 we get that,

• ${x}^{2} + (a + 1)x + b = 0$

• $( {2}^{2} ) + (a + 1)2 + b = 0$

• $4 + 2a + 2 + b = 0$

• $6 + 2a + b = 0 - - - - - (i)$

$\mathcal\purple{♧Next,}$

• Taking x as -3 in applying the given quadratic polynomial we get that,

• ${x}^{2} + (a + 1)x + b = 0$

• $({ - 3}^{2} ) + (a + 1)( - 3) + b = 0$

• $9 - 3a - b + b = 0$

$6 - 3a + b = 0 - - - - - (ii)$

• Here we compare these two equations we get that,

• $6 + 2a + b = 6 - 3a + b$

• $6 - 6 + 2a + 3a + b - b = 0$

• $5a = 0 = a = 0$

$\mathcal\pink{♧Now}$

• Substituting the value if a =0 in ii equation we get that,

• $6 - 3a + b = 0$
• $6 - 3(0) + b = 0$

• $6 - 0 - b = 0$

• $b = - 6$

$\mathcal\purple{♧Therefore}$

• The value of a =0
• The value of b=-6.

$\mathcal\purple{Option D is the answer}$

♧Hope it helps u mate .

♧Thank you .