Question

The coefficient of x in the eq is
${x}^{2} + px + q = 0$
was wrongly written as 17 in the place of 13 and the roots formed were -2 and -15. The roots of the correct eq. are ?

Hint - we can do it by product and sum of roots, but I m unable to to...plz help

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Answer 1

Answer:

x = -3 or x = -10

Step-by-step explanation:

GIVEN =>

coefficient of x² = 1

coefficient of x = 17 ( taken wrongly )

constant = q

roots of equation = - 2 , - 15

=> we know , product of roots = constant/coefficient of x²

=> -2 × -15 = q / 1

=> 30 = q

now , the equation becomes =>

x² + 13 x + 30 = 0

(given that , correct value of p = 13 )

=> x² + 13 x + 30 = 0

=> x² + 10x + 3x + 30 = 0

=> x ( x + 10 ) + 3 ( x + 10 ) = 0

=> ( x + 10 ) ( x + 3 ) = 0

so , either ( x + 10 ) = 0

=> x = -10

or ( x + 3 ) = 0

=> x = -3