Question

If 2 and 1 are roots of quadratic equation ax^2+bx+2=0 , then find value of a and b

Answer 1

$\large\bf\underline{Given:-}$

• 2 and 1 are the roots of the quadratic equation ac² + bx + 2 = 0

$\large\bf\underline {To \: find:-}$

• Value a and b

$\huge\bf\underline{Solution:-}$

roots = 2 and 1

• so, x = 2 or x = 1

★ p(x) = ax² + bx + 2

putting value of x = 2 in the given quadratic equation.

$\dashrightarrow \rm \: a {x}^{2} + bx + 2 = 0 \\ \\ \dashrightarrow \rm \: a \times {2}^{2} + b \times 2 + 2 = 0 \\ \\ \dashrightarrow \rm \: 4a + 2b + 2 = 0 \\ \\ \rm \dag \: divide \: both \: side \: by \: 2 \\ \\ \dashrightarrow \rm \: 2a + b = - 1......(i)$

putting x = 1 in the given quadratic equation.

$\dashrightarrow \rm \: {ax}^{2} + bx + 2 = 0 \\ \\ \dashrightarrow \rm \: a \times {1}^{2} + b \times 1 + 2 = 0 \\ \\ \dashrightarrow \rm \: a + b = - 2........(ii)$

From eq. (i) and (ii) .

$\rm \: 2a + b = - 1 \\ \rm \: \: \: a + b = - 2 \\ \: \: \: - \: \: \: - \: \: \: \: \: + \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: \: a \: \: \: \: \: \: \: \: = 1 \\ \\ \rm \dag \: putting \: value \: of \: a = 1 \: in \: eq.(i) \\ \\ \rm \rightarrowtail \: 2a + b = - 1 \\ \\ \rm \rightarrowtail2 \times 1 + b = - 1 \\ \\ \rm \rightarrowtail2 + b = - 1 \\ \\ \rm \rightarrowtail \: b = - 1 - 2 \\ \\ \rm \rightarrowtail \: b = - 3$

So,

$\bf \star \: a = 1 \: and \: b = - 3$

$\underline{\bf\bigstar \: Verification:-}$

$\longmapsto \rm \: p(x) = a {x}^{2} + bx + 2 = 0 \\ \\ \rm \: \: putting \: value \: of \: a \: and \: b \\ \\ \longmapsto \rm \: p(x) = {x}^{2} - 3x + 2 \\ \\ \longmapsto \rm \: {x}^{2} - x - 2x + 2 \\ \\ \longmapsto \rm \: x(x - 1) - 2(x - 1) \\ \\ \longmapsto \rm \: (x - 2)(x - 1) \\ \\ \longmapsto \bf\: x = 1 \: or \: x \: = 2$

So, we get the same zeroes 2 and 1 that are mentioned in the Question.

hence Verified

Answer 2

Given :

The polynomial is a(x)² + bx + 2 and their roots are 2 and 1

To Find :

The value of a and b

Solution :

Since , 2 and 1 are the roots of given polynomial

Thus ,

a(2)² + b(2) + 2 = 0

4a + 2b + 2 = 0

2a + b + 1 = 0

2a + b = -1 ----- (i)

and

a(1)² + b(1) + 2 = 0

a + b + 2 = 0

a + b = -2 ------- (ii)

Multiply eq (ii) by 2 , we get

2a + 2b = -4

Subtract eq (i) from eq (ii) , we get

2a + 2b - (2a + b) = -4 - (-1)

2b - b = -3

b = -3

Put the value of b = -3 in eq (ii) , we get

a + (-3) = -2

a = -2 + 3

a = 1

$\therefore \sf \underline{The \: value \: of \: a \: and \: b \: are \: 1 \: a nd -3}$