Question

lf one zero of the polynomial p(x)=ax square-3(a-1)x-1 is 1 . find the value of a.​

Answer 1

SOLUTION:-

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putting the value of x=1 inp(x) = 0

x=1,must satisfy the in given equation

ax²-3(a-1)x-1

hence, a*1²-3(a-1)1-1 =0

or, a-3a+3-1=0

or, -2a=-2

or a= 1

Hope you would like it.....

Answer 2

$HEYA \: \\ \\for \: a \: general \: quadratic \: \\ equation \: say \: \\ p(x) = ax {}^{2} + bx + c \\ \\ \alpha + \beta = - b \div a \\ and \\ \alpha \beta = c \div a \\ \\ where \: \alpha \: and \: \beta are \: its \: two \: roots \: \\ \\ p(x) = ax {}^{2} - 3(a - 1)x - 1 \\ \\ one \: \: root \: is \: given \: as \: \: 1 \: \\ let \: the \: other \: root \: be \: \alpha \\ \\ \alpha + 1 =( 3(a - 1)) \div a ...Equation \: (i) \: \\ and \\ \alpha \times 1 = - 1 \div a \\ \\ \alpha = - 1 \div a ...put \: this \: value \: of \: \\ \alpha \: in \: above \: equation \: we \: have \: \\ \\ (a - 1) = 3(a - 1) \\ \\ a - 1 - 3a + 3 = 0 \\ \\ - 2a + 2 = 0 \\ \\ - 2a = - 2 \\ \\ a = 1$