Question

4x+6/x+13=0 your answer correct to two decimal places by quadratic equation ​

Answer 1

SolutioN :

We have, Quadratic Equation.

$\tt : \implies 4x + \dfrac{6}{x} + 13 =0 .$

$\tt : \implies \dfrac{4 {x}^{2} + 6 + 13x}{x} =0 .$

$\tt : \implies 4 {x}^{2} + 6 + 13x=0 \times x .$

$\tt : \implies 4 {x}^{2} + 6 + 13x=0.$

$\tt : \implies 4 {x}^{2} + 13x + 6 =0 .$

★ Compare with General Formula.

$\tt \dagger \: \: \: \: \: a {x}^{2} + bx + c=0.$

Where as,

• a = 4.
• b = 13.
• c = 6.

Now, By Quadratic Formula.

$\tt \dagger \: \: \: \: \: \fbox{ x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\tt : \implies x = \frac{ - 13 \pm \sqrt{ { \big(13 \big)}^{2} - 4 \times 4 \times 6 } }{2 \times 4}$

$\tt : \implies x = \dfrac{ - 13 \pm \sqrt{ 169 - 96 } }{8}$

$\tt : \implies x = \dfrac{ - 13 \pm \sqrt{ 73 } }{8}$

Therefore, the value of x = 13 ± √73 / 8.

Answer 2

$\large\underline{\underline{\mathbb{\purple{ANSWER}}}}$

$\longrightarrow \tt{4x +\dfrac{6}{x}+ 13 = 0 } \\\\ \longrightarrow \tt{\dfrac{4x^2 + 6 + 13x}{x}= 0} \\\\\longrightarrow \tt{4x^2 + 13x + 6 = 0}\\\\ \rm{Now , we \;got \;a\; quadratic\; equation }\\\\\rm{Finding \;its\; roots \;by\; using \;quadratic\; formula}\\\\ \dagger\boxed{\bf{x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}}}$

Where

• b = 13
• a = 4
• c = 6

Substituting the values we have

$\longrightarrow \tt{x = \dfrac{-13 \pm \sqrt{13^2 - 4(4)(6)}}{2(4)}} \\\\ \longrightarrow \tt{\dfrac{-13 \pm \sqrt{169 - 96 }}{8}}\\\\ \longrightarrow \tt{x = \frac{-13+\sqrt{73}}{8}(or) \frac{-13-\sqrt{73}}{8}}$

$\therefore\underline\orange{{\sf{Hence,\; x =\dfrac{ -13\pm\sqrt{73}}{8}}}}$