Use the discriminant to determine the number of real roots the equation has.
3x2 – 5x + 1 =0
(a) One real root (a double root),
(b) Two distinct real roots,
(c) Three real roots,
(d) None (two imaginary roots)
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Discriminant = bx2 – 4ac
Compare the above equation 3x2 – 5x + 1 =0 with ax2 + bx + c = 0
We get, a = 3, b = -5, c = 1
Put the value of a, b and c;
Discriminant = bx2 – 4ac
Discriminant = (-5)2 - 4 × 3 × 1
= 25 – 12
= 13 [13 > 0]
Therefore, discriminant is positive.
So the given equation has two distinct real roots.
Answer - B
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✌️_______UR ANSWER_______✌️
Discriminant = bx2 – 4ac
Compare the above equation 3x2 – 5x + 1 =0 with ax2 + bx + c = 0
We get, a = 3, b = -5, c = 1
Put the value of a, b and c;
Discriminant = bx2 – 4ac
Discriminant = (-5)2 - 4 × 3 × 1
= 25 – 12
= 13 [13 > 0]
Therefore, discriminant is positive.
So the given equation has two distinct real roots.
Answer - B
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6
" b "
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Anonymous:
by the way aapne shi pkra hai
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