Question

If the zeroes of the polynomial x^2 - ax + b are 3 & 4, then a & b are respectively equal to.... explain ans be star⭐

Answer 1

According to question
${x}^{2} - ax + b$
3 & 4 the zeroes of the polynomial
First equation
$( {3}^{2} ) - a(3) + b = 0 \\ 9 - 3a + b = 0 \\ b = 3a - 9$
second equation
$( {4}^{2} ) - a(4) + b = 0 \\ 16 - 4a + b = 0 \\ 16 + b = 4a \\ \frac{16 +b }{4} = a$
putting the value of a in equation i
$b = 3a - 9 \\b = 3( \frac{16 + b}{4}) - 9 \\ b = \frac{48 + \: 3b - 36}{4} \\ cross \: multiply \\ 4b = 12 + 3b \\ 4b - 3b = 12 \\ b = 12$
putting the value of b in equation ii
$\frac{16 + b}{4} = a \\ \frac{16 + 12}{4} = a \\ \frac{28}{4} = a \\ \: \: 7 = a$

Answer 2

Step-by-step explanation:

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