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For what value(s) of ‘a’ quadratic …

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For what value(s) of ‘a’ quadratic equation 30ax^2-6x+1=0 has no real roots?

Posted by The_ Pj 4 years, 8 months ago

CBSE > Class 10 > Mathematics

4 answers

Simranpreet Kaur 4 years, 8 months ago

The equation has no real root i. e. B2-4AC <0 here b=-6, a=3a and c=1 (-6*-6) -4(3a)(1) <0 36-12a<0 36<12a 36/12<a 3<a

Kumar Kumar 4 years, 8 months ago

Here,coefficients of standard form of equation Ax²+Bx+C=0 are A=30a B=-6 C=1 So,Discriminant,D=B²-4AC Since,equation has no real roots i.e. D<0 i.e. B²-4AC<0 i.e. (-6)² -4 (30a)(1)<0 i.e. 36-120a <0 On transposing (-120a) i.e. 36 <0-(-120a) i.e. 36 <120a i.e. 6×6 <6×(20a) On cancelling out 6 from both sides i.e. 6<20a On dividing both sides by 20 i.e. 6/20 < a i.e. 3/10 < a i.e. a >3/10 Hence, a>3/10 for which 30ax² +6x+1=0 gas no real roots

Kumar Kumar 4 years, 8 months ago

B= ±6 as B²=36

Kumar Kumar 4 years, 8 months ago

Here,coefficients of standard form of equation Ax²+Bx+C=0 are A=30a B=6 C=1 So,Discriminant,D=B²-4AC Since,equation has no real roots i.e. D<0 i.e. B²-4AC<0 i.e. (6)² -4 (30a)(1)<0 i.e. 36-120a <0 On transposing (-120a) i.e. 36 <0-(-120a) i.e. 36 <120a i.e. 6×6 <6×(20a) On cancelling out 6 from both sides i.e. 6<20a On dividing both sides by 20 i.e. 6/20 < a i.e. 3/10 < a i.e. a >3/10 Hence, a>3/10 for which 30ax² +6x+1=0 gas no real roots

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