Question

Equation x^2 + px + q = 0 where p and q are constants, has roots -3 and 5
1). find p and q

Answer 1

Answer:

$\green {Value \: of \: p = -2 \:and \: q=-15}$

Step-by-step explanation:

$Given \: -3\:and \: 5 \:are \: roots \: of \\quadratic \: equation \: x^{2}+pc+q=0 ,$

$Substitute \:\pink{ x= -3}\:in \: the \\ equation ,we \:get$

$(-3)^{2} + p(-3) + q = 0$

$\implies 9 - 3p + q = 0$

$\implies q = 3p - 9 \: ---(1)$

$Substitute \:\pink{ x= 5}\:in \: the \\ equation ,we \:get$

$5^{2} + p \times 5 + q = 0$

$\implies 25 + 5p + q = 0$

$\implies q = -5p - 25 \: ----(2)$

$\orange {(1) = (2)}$

$3p - 9 = -5p - 25$

$\implies 3p + 5p = 9 - 25$

$\implies 8p = - 16$

$\implies p = \frac{-16}{8} = -2$

$Substitute \:\pink{ p = -2}\:in \: the \\ equation \:(1),we \:get$

$\implies q = 3\times (-2)- 9$

$= -6 -9 = -15$

Therefore.,

$\green {Value \: of \: p = -2 \:and \: q= -15}$

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