Question

find the roots of the quadratic equation by applying the quadratic formula :- (i) x/x+1 + x+1/x = 34/ 15 , x is not equal to -1,0​

Answer 1

$\mathsf{ \frac{x}{x + 1} + \frac{x + 1}{x} = \frac{34}{15} } \\ \\ \mathsf{ \frac{ {x}^{2} + {(x + 1)}^{2} }{x(x + 1)} = \frac{34}{15} } \\ \\ \mathsf{ \frac{ {x}^{2} + {x}^{2} + 2x + 1}{ {x}^{2} + x } = \frac{34}{15} } \\ \\ \mathsf{15( 2{x}^{2} + 2x + 1) = 34( {x}^{2} + x )} \\ \\ \mathsf{30 {x}^{2} + 30x + 15 = 34 {x}^{2} + 34x} \\ \\ \mathsf{34 {x}^{2} - 30 {x}^{2} + 34x - 30x - 15 = 0 } \\ \\ \mathsf{4 {x}^{2} + 4x - 15 = 0} \\ \\ \mathsf \red{solving \: by \: using \: quadratic \: formula \: : } \\ \\ \boxed{ \mathsf \blue{x \: = \: \frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac } }{2a} }} \\ \\ \mathsf{here \: \: \: \: a \: = \: 4 \: \: \: \: \: b \: = \: 4 \: \: \: \: \: c \: = \: - 15} \\ \\ \mathsf{x = \frac{ - (4) \: \pm \: \sqrt{ {4}^{2} - 4(4)( - 15) } }{2 \times 4} } \\ \\ \mathsf{x = \frac{ - 4 \: \pm \: \sqrt{16 + 240} }{8} } \\ \\ \mathsf{x = \frac{ - 4 \: \pm \: \sqrt{256} }{8} } \\ \\ \mathsf{x = \frac{ - 4 \: \pm \: 16}{8} } \\ \\ \mathsf{x = \frac{ - 4 + 16}{8} \: ; \: \frac{ - 4 - 16}{8} } \\ \\ \mathsf{x = \frac{12}{8} \: ; \: \frac{ - 20}{8} } \\ \\ \mathsf{x \: = \: \frac{3}{2} \: ; \: \frac{-5}{2}}\\ \\ \boxed{\mathsf{ \therefore \: roots \: are \: \: \: x = \frac{3}{2} \:; \: \frac{ - 5}{2} }}$

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