Question

यदि वर्ग समीकरण 9x²-kx+16=0 के मूल समान है तो k का मान ज्ञात कीजिए

Answer 1

b^2-4ac=0
(-k)^2-4*9*16=0
K=24

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The value of k when the roots of the quadratic equation 9x²-kx+16=0 are equal

FORMULA TO BE IMPLEMENTED

Two roots of the quadratic equation

$\sf{a {x}^{2} + b x+ c = 0\: }$

will be equal if

$\sf{Discriminant= D = {b}^{2} - 4ac = 0 }$

CALCULATION

The given Quadratic Equation is

$\sf{9 {x}^{2} - kx + 16 = 0 \: }$

$\sf{comparing \: with \: \: a {x}^{2} + b x+ c = 0\: } \: \: we \: get$

$\sf{a = 9 \: , \: b = - k \: , \: c= 16}$

$\therefore \: \sf{Discriminant= D = {b}^{2} - 4ac }$

$= \sf{ {( - k)}^{2} - 4 \times 9 \times 16 \: }$

$= \sf{ { k}^{2} - 4 \times 9 \times 16 \: }$

Since the roots are equal

$\sf{ { k}^{2} - 4 \times 9 \times 16 = 0\: }$

$\implies \: \sf{ { k}^{2} = 4 \times 9 \times 16 \: }$

$\implies \: \sf{ { k} = \pm \: ( 2 \times 3\times 4) \: }$

$\implies \: \sf{ { k} = \pm \: \: 24\: }$

RESULT

$\boxed{ \sf{ \: k \: = \pm \: \: 24 \: \: }}$

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