Math, asked by 1712003, 1 year ago

find the condition for which the roots of the equation ax²-(a+1)x+1=0 are always available

Answers

Answered by shadow1924
1
It's delta should always be greater than or equal to zero
i.e.
 {b}^{2}  - 4ac \geqslant 0
 { (- a - 1)}^{2}  - 4 \times a \times 1 \geqslant 0
 {a}^{2}  + 2a + 1 - 4a \geqslant 0
 {(a - 1)}^{2}  \geqslant 0
therefore
a \geqslant 1

1712003: thanks
shadow1924: welcome
1712003: is the value of a real?
shadow1924: yes..its greater than 0 always
1712003: okk i got
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