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The range of values of a for which roots of the quadratic equation x^2+x-3a+2=0 are of opposite sign is given by a) a b) a>2 c) 1 d) 0
Your given equation is x2 + x – 3a + 2 = 0 Since the roots are opposite in sign, so α x β is –ve (i) Hence, α x β = -3a + 2/1 (ii) Therefore, product of roots of the eq. ax2 + bx + c = c/a From equation (i) and (ii), we have -3a + 2 < 0 -3a < -2 a > 2/3 (i) Also, discriminant D should be greater than 0 for roots to be real Hence, 1 – 4 (-3a + 2) > 0 1 > 4(-3a + 2) -12a + 8 < 1 -12a < -7 a > 7/12 (ii) Taking the intersection (i) and (ii) we have, a > 2/3 Hence, the most appropriate option is (b) a > 2 , since a > 2/3 Then, a > 2.
Your given equation is
x2 + x – 3a + 2 = 0
Since the roots are opposite in sign, so α x β is –ve (i)
Hence,
α x β = -3a + 2/1 (ii)
Therefore, product of roots of the eq.
ax2 + bx + c = c/a
From equation (i) and (ii), we have
-3a + 2 < 0
-3a < -2
a > 2/3 (i)
Also, discriminant D should be greater than 0 for roots to be real
Hence, 1 – 4 (-3a + 2) > 0
1 > 4(-3a + 2)
-12a + 8 < 1
-12a < -7
a > 7/12 (ii)
Taking the intersection (i) and (ii) we have,
a > 2/3
Hence, the most appropriate option is (b) a > 2 , since a > 2/3
Then, a > 2.
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