Question -
Answer -
(i) 2x2┬атАУ3x┬а+ 5 = 0
(ii) 3x2┬атАУ┬а4тИЪ3x┬а+ 4 = 0
(iii) 2x2┬атАУ┬а6x┬а+ 3 = 0
Solution
(i) Given,
2x2┬атАУ 3x┬а+ 5 = 0
Comparing the equation with┬аax2┬а+┬аbx┬а+┬аc┬а=0, we get
a┬а= 2,┬аb┬а=-3 and┬аc┬а= 5
We know, Discriminant =┬аb2┬атАУ4ac
=┬а(тАУ 3)2┬атАУ 4 (2) (5) = 9 тАУ 40
= тАУ 31
As you can see, b2┬атАУ 4ac <0
Therefore, no real root is possible for thegiven equation,┬а2x2┬атАУ 3x┬а+ 5 = 0.
Solution
(ii) 3x2┬атАУ 4тИЪ3x┬а+4 = 0
Comparing the equation with┬аax2┬а+┬аbx┬а+┬аc┬а=0, we get
a┬а= 3,┬аb┬а=┬а-4тИЪ3┬аand┬аc┬а=4
We know, Discriminant =┬аb2┬атАУ4ac
= (-4тИЪ3)2┬атАУ 4(3)(4)
= 48 тАУ 48 = 0
As┬аb2┬атАУ 4ac┬а=0,
Real roots exist for the given equation andthey are equal to each other.
Hence the roots will be тАУb/2a┬аand┬атАУb/2a.
тАУb/2a┬а= -(-4тИЪ3)/2├Ч3 = 4тИЪ3/6= 2тИЪ3/3 = 2/тИЪ3
Therefore, the roots are┬а2/тИЪ3┬аand2/тИЪ3.
Solution
(iii) 2x2┬атАУ┬а6x┬а+3 = 0
Comparing the equation with┬аax2┬а+┬аbx┬а+┬аc┬а=0, we get
a┬а= 2,┬аb┬а=-6,┬аc┬а= 3
As we know, Discriminant =┬аb2┬атАУ4ac
= (-6)2┬атАУ 4 (2) (3)
= 36 тАУ 24 = 12
As┬аb2┬атАУ 4ac┬а>0,
Therefore, there are distinct real roots existfor this equation, 2x2┬атАУ┬а6x┬а+ 3 = 0.
= (-(-6) ┬▒ тИЪ(-62-4(2)(3)) )/ 2(2)
= (6┬▒2тИЪ3 )/4
= (3┬▒тИЪ3)/2
Therefore the roots for the given equation are(3+тИЪ3)/2 and (3-тИЪ3)/2