Question

10. If a and Bare the roots of equation ax^2 - bx + c =0, then find the value of a + B.(1)​

Answer 1

Step-by-step explanation:

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

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.

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

putting x = 1 in the given quadratic equation.

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

From eq. (i) and (ii) .

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

So,

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

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

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

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

✵Hence Verified✵