Question

The quadratic equation whose roots are 7 + √3 and 7 - √3 is
a) x2 + 14x - 46 = 0
b) x2 - 14x + 46 = 0
c) x2 - 14x - 46 = 0
d) x2 + 14x + 46 = 0​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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GiveN :

$\rightsquigarrow \sf \: Two \: \: roots \: \: are \\ \sf1) \: 7 + \sqrt{3} \\ \sf2) \: 7 - \sqrt{3}$

To FinD :

The correct polynomial,

$\begin{array}{c} \: \sf \: a) \: {x}^{2} + 14x - 46 \\ \sf \: b) {x}^{2} - 14x + 46 \\ \sf \: c) \: {x}^{2} - 14x - 46 \\ \sf \: d) \: {x}^{2} + 14x + 46 \end{array}$

SolutioN :

$\circ \: \: \sf \: we \: \: know, \: for \: \: any \: \: quadratic \: \: polynomial$

$\: \: \: \sf \: \: {x}^{2} - (sum \: \: of \: \: roots)x + \: product \: \: of \: \: roots = 0$

Hence, the eqn will be

$\dashrightarrow \sf \: {x}^{2} - (7 + \sqrt{3} + 7 - \sqrt{3} )x + (7 + \sqrt{3} )(7 - \sqrt{3} ) = 0$

$\dashrightarrow \sf \: {x}^{2} - ( 14 )x + 46 = 0$

$\: \therefore \: \: \: \: \underline{ \boxed{ \sf{ b) \: \: {x}^{2} - 14x + 46 = 0}}}$

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