Math, asked by NimmiSoni, 11 months ago

16 X square - 24 x minus 13 is equal to zero find the roots of the following quadratic equation​


Varun1870: Hi
Anonymous: ___k off

Answers

Answered by TRISHNADEVI
13

  \red{ \huge{ \underline{ \overline{ \mid{ \bold{ \purple{ \:  \: SOLUTION \:  \:   \red{ \mid}}}}}}}}

 \underline{ \bold{ \:  \:GIVEN \:  \:  :}} \to \:   \\  \\  \bold{The \:  \: quadratic\:  \: equation \:  \: is \:  : } \\  \\  \bold{16x {}^{2} - 24x - 13 = 0 }

 \underline{ \bold{ \: We \:  \: know \:  \: that \:  \: }} \\  \\  \bold{ A \:  \: quadratic \:  \: equation \:  \: is \: \: in \:   \: the \:  \: form} \\  \\  \bold{of \:  \:  \: ax {}^{2}  + bx + c = 0 \:  \: and \:  \: its \:  \: roots \:  \: are \:  : } \\  \\  \bold{x =  \frac{  \:  \:  \: - b \:  \pm \:  \sqrt{ b {}^{2}  - 4ac} \:  \:  \:  \:  \: }{2a} }

 \bold{Here,} \\  \\   \:  \:  \:  \:  \:  \:  \:  \: \bold{a = 16} \\  \\  \:  \:  \:  \:  \:  \:  \:  \: \bold{b =  - 24}  \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \bold{c =  - 13}

 \bold{ \therefore \:  \: x =  \frac{  \:  \: - b \:  \pm \:  \sqrt{b {}^{2}  - 4ac}  \:  \:  \: }{2a} } \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ =  \:  \:  \frac{ - ( - 24) \:  \pm \:   \sqrt{( - 24) {}^{2} \:   -  \: 4 \times 16 \times ( - 13) \:  \: } }{2 \times 16} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ = \frac{ \:  \: 24 \:  \pm \:  \: \sqrt{576 + 832}   \:  \: }{32}  } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ =  \frac{24 \:  \pm \:  \sqrt{1408} \:  \:  }{32} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ =  \frac{ \:  \: 24 \:  \pm \: 8  \sqrt{22}  \:  \: } {32} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ =  \frac{ \cancel{8} \: (3 \:  \pm \:  \sqrt{22} ) \:  \:  \: }{ \cancel{8} \times 4} } \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \bold{ =  \frac{ \:  \: 3 \:  \pm \:  \sqrt{22} \:   \: }{4} }

    \bold{  \therefore \:  \: x=  \frac{ \:  \: 3 +  \sqrt{22} \:  \:  }{4}  \:  \:  \:  \: Or ,\:  \:  \:  \: x =   \frac{ \:  \: 3 -  \sqrt{22}  \:  \: }{4} }

  \bold{ Hence, \:  \:  \: the \:  \: roots \:  \: of \:  \: the \:  \: quadratic \:  \: equation } \\  \\   \bold{ 16x {}^{2}   - 24x - 13 = 0 \:  \:  \: are \:  \:  : } \\  \\  \red{ \bold{ \:  \: x =  \frac{ \:  \: 3 +  \sqrt{22}  \:  \: }{4}  \:  \:  \:  \:  \:  \:  ,\:  \:  \:  \:  \:  \frac{ \:  \: 3 -  \sqrt{22}  \:  \: }{4} }}


AvishantRoy: how u typied like this
TRISHNADEVI: With Latex
Answered by AvishantRoy
1

16x^2-24x-13=0

using completing square method

16x^2-24x=13

(4x)^2-2×4x×3=13

adding 3^2 on both sides

(4x)^2-2×4x×3+3^2=13+3^2

(4x-3)^2=22

4x-3=+-√22

x=(3+-√22)/4

hence two roots of given eq. are ( 3+√22)/4 and ( 3-√22)/4

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