1 thought on “The quadratic graph whose sum of zeroes is 0 and that of its product is -1 is given by <br /><br />(a) y = x2<br /><br />– x (b) y”
Step-by-step explanation:
[tex]given \: that \: sum \: of \: zeroes \: is \: 0 \\ \: and \: product \: is \: – 1[/tex]
we know that sum of zerois -b/a and product of the zeroes is c/a
[tex] \frac{0}{1} = \frac{ – b}{a} \\ here \: we \: got \: b \: = 0 \: and \: a = 1 \: [/tex]
nowlet’s find c
[tex] \frac{ – 1}{1} = \frac{c}{a} \\ here \: from \: the \: sum \: of \: the \: equation \: \\ we \: got \: that \: a \: = 1 \: and \: c \: = – 1[/tex]
Step-by-step explanation:
[tex]given \: that \: sum \: of \: zeroes \: is \: 0 \\ \: and \: product \: is \: – 1[/tex]
we know that sum of zero is -b/a and product of the zeroes is c/a
[tex] \frac{0}{1} = \frac{ – b}{a} \\ here \: we \: got \: b \: = 0 \: and \: a = 1 \: [/tex]
now let’s find c
[tex] \frac{ – 1}{1} = \frac{c}{a} \\ here \: from \: the \: sum \: of \: the \: equation \: \\ we \: got \: that \: a \: = 1 \: and \: c \: = – 1[/tex]
therefore we got all the values
[tex]quadratic \: equation \: is \\ {ax}^{2} + bx + c = 0 \\ 1 {x}^{2} + 0x + ( – 1) = 0 \\ therefore \: we \: got \\ {x}^{2} + 0x – 1 = 0[/tex]
Hope it helps