Question

if a and b are the roots of 4x^2+3x+7=0 then the value of 1/a+1/b is

Answer 1

Hi there

Given :-

$4 {x}^{2} + 3x + 7 = 0$
Where


$\alpha = 4 \\ \\ \beta = 3 \\ \\ \gamma = 7 \\ and \: also \: \\ \\ a + b \: \: \: = \frac{ - \beta }{ \alpha } \\ \\ = > a + b \: \: = \frac{ - 3}{4} \\ \\ and \: ab \: = \frac{ \gamma }{ \alpha } = \frac{7}{4}$
so the value of
$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \\ \\ = > \frac{1}{a} + \frac{1}{b} \: = \frac{ - 3}{4} \div \frac{7}{4} \\ \\ = \frac{ - 3}{4} \times \frac{4}{7} \\ \\ = \frac{ - 3}{7}$
Hope this helps ya☺

Answer 2

Answer: $\frac{-3}{7}$

Step-by-step explanation:

• The roots of quadratic equation $4x^{2} +3x+7=0$ must be calculated.
• a=4,b=3 and c=7.
• D=$\sqrt{b^{2}-4ac }$

          =$\sqrt{3^{2}-4*4*7 }$

          =$\sqrt{9-112}$

          =$\sqrt{-103}$

          =$\sqrt{103} \ i$       [∵ $i^{2}=-1$]

• So,the roots of the equation are

            $\frac{-b+D}{2a}$ and  $\frac{-b-D}{2a}$

       ⇒$\frac{-3+\sqrt{103} i }{8}$ and $\frac{-3-\sqrt{103}i }{8}$

• Now,$\frac{1}{a}+\frac{1}{b}$

             ⇒ $\frac{8 }{-3+\sqrt{103} i}+\frac{8 }{-3-\sqrt{103} i}$

             ⇒$\frac{-24-8\sqrt{103}i+(-24+8\sqrt{103}i ) }{(-3)^{2} -( \sqrt{103} i)^{2} }$

             ⇒$\frac{-24-8\sqrt{103}i-24+8\sqrt{103}i ) }{9 +{103} }$

             ⇒$\frac{-48 }{112 }$

             ⇒$\frac{-3}{7}$

• Hence,the required sum of the reciprocal of roots is $\frac{-3}{7}$.

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