Find the roots of the quadratic equations given in Q. 1, above b

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28. Find the roots of the quadratic equations given in Q. 1, above by applying the quadratic formula.


(i) We have,
2x2 – 1x + 3 = 0
Here, a = 2, b = –7 and c = 3
Now, b2 – = (–7)2 – 4(2) (3)
= 49 – 24 = 25 Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 – 1x + 3 = 0 has distinct roots i.e. α and β.
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#7 {main}</pre>
equals fraction numerator negative left parenthesis negative 7 right parenthesis plus-or-minus square root of 25 over denominator 2 cross times 2 end fraction equals fraction numerator 7 plus-or-minus 5 over denominator 4 end fraction
So,             straight alpha space equals space fraction numerator 7 plus 5 over denominator 4 end fraction equals 12 over 4 equals 3
straight beta space equals space fraction numerator 7 minus 5 over denominator 4 end fraction equals 2 over 4 equals 1 half
Therefore, the required roots are 3 and 1 half.

(ii) We have,
2x2 + x – 4 = 0
Here, a = 2, b = 1 and c = –4
Now, b– 4ac = (1)2 –4 (2) (–4)
= 1 + 32 = 33
Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 + x – 4 = 0 has distinct roots i.e. α and 946;.
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                  equals fraction numerator negative 1 plus-or-minus square root of 33 over denominator 2 left parenthesis 2 right parenthesis end fraction equals fraction numerator negative 1 plus-or-minus square root of 33 over denominator 4 end fraction

So,           straight alpha equals fraction numerator negative 1 plus square root of 33 over denominator 4 end fraction
                straight beta equals fraction numerator negative 1 minus square root of 33 over denominator 4 end fraction

(iii) We have,
         4 straight x squared plus 4 square root of 3 straight x plus 3 equals 0
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Now,     straight b squared minus 4 ac space equals space left parenthesis 4 square root of 3 right parenthesis squared minus 4 space left parenthesis 4 right parenthesis space left parenthesis 3 right parenthesis
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#6 {main}</pre>
Therefore, the quadratic equation 4 straight x squared plus 4 square root of 3 straight x plus 3 equals 0  has equal roots i.e. straight alpha equals straight beta
Now,         space straight x equals fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction
space space space space equals space fraction numerator negative 4 square root of 3 plus-or-minus 0 over denominator 2 cross times 4 end fraction equals fraction numerator negative 4 square root of 3 over denominator 8 end fraction
     Therefore, the requried roots are equals fraction numerator negative square root of 3 over denominator 2 end fraction. 

(iv) We have,
2x2 + x + 4 = 0
Here, a = 2, b = 1 and c = 4
Now, b– 4ac = (1)2 – 4(2) (4)
= 1 – 32 = – 31
Since, b2 – 4ac = 0
Therefore, the quadratic equation 2x2 + x + 4 = 0 has no roots.

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