State the nature of roots of quadratic equation 9x^2=_1
sofiya:
It refers to 9x square
By the way I got the answer
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9x^2 +1 =0 comparing it with ax^2 +bx +c = 0 we get a= 9 , b= 0 , c = 1 now
D = ✔️b^2-4ac = ✔️0-4x9x1 = ✔️-36 => roots are imaginary
Which type of roots is this?
When we get a minus sign in the root it shows that roots are imaginary
Mark as best Plz
anyway there is no use of taking root here. simply we can solve here by taking discriminent
Gud thank you
9 x^2 = -1 = i^2 ;; as i = root(-1). x = + 1/3 i or -1/3 i.. imaginary
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9x^2 +1 =0 , here a=9, b=0, c=1
then, discriminent Δ = b² - 4ac
Δ = 0² - 4×9×1
Δ = -36
here the result obtained is less than 0. so, the roots are imaginary.
then, discriminent Δ = b² - 4ac
Δ = 0² - 4×9×1
Δ = -36
here the result obtained is less than 0. so, the roots are imaginary.
easily understandable. nice
9 x^2 = -1 = i^2 ;; as i = root(-1). x = + 1/3 i or -1/3 i.. imaginary number (0,1/3) or (0,-1/3)
It was really helpful
Thank you so much
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