Question

if 2 and -3 are zeros of
${x}^{2} + (a + 1)x + b$
then find the value of a and b​

Answer 1

Answer :

The value of a = 0 and b = -6

Given :

The quadratic polynomial is

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

To Find :

The value of a and b

Solution :

We know that ,

$\sf{sum \: of \: the \: zeroes = \dfrac{ - coefficient \: of \: x }{coefficient \: of \: {x}^{2} } }$

$\sf2 + ( - 3) = \dfrac{ -( a + 1)}{1} \\ \implies \sf2 - 3 = - a - 1 \\ \implies \sf - 1 = - a - 1 \\ \sf \implies a + 1 = 1 \\ \implies \boxed{ \sf{a = 0} }$

And again ,

$\sf{product \: of \: the \: zeroes = \dfrac{constant \: \: term}{coefficient \: of \: {x}^{2} } }$

$\sf(2) \times ( - 3) = \frac{b}{1} \\ \implies \sf - 6 = b \\ \implies \boxed{ \sf{b = - 6}}$

Thus the quadratic polynomial obtained by putting the values of a and b is :

$\sf {x}^{2} + (0 + 1)x + ( - 6) \\ = \sf {x}^{2} + x - 6$

Answer 2

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QUESTION :

if 2 and -3 are zeros of

${x}^{2} + (a + 1)x + b$

then find the value of a and b.

SOLUTION :

From the above Question, we can gather the following information......

The given Zeroes are 2 and -3

Therefore,

Sum of Zeroes = 2 + ( -3 ) = -1

Product of Zeroes = 2 × (-3 ) = -6

Now the given polynomial :

${x}^{2} + (a + 1)x + b$

We know what :

$Sum \: of \: Zeroes = \dfrac{-b}{a} = 1 - a$

$So \: 1 - a = -1 \\ \\ a = 2$

We know that :

$Product \: Of \: Zeroes = \dfrac{c}{a} = b = 6$

$So \: b = 6$

Hence The values of a and b are 2 and 6 respectively.

ANSWER :

The values of a and b are 2 and 6 respectively.