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Standard XII

Mathematics

nth Term of A.P

Two candidate...

Question

Two candidates attempt to solve a quadratic of the form $x^{2} + p x + q = 0$ P was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original equation.

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Solution

$x^{2} + p x + q = 0$
$p = 13$
$q = - 2 \times (- 15) = 30$
$x^{2} + 13 x + 30 = (x + 10) (x + 3)$
so the roots are $- 3, - 10$

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Two candidates attempt to solve a quadratic of the form x2+px+q=0 P was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original equation.

x2+px+q=0p=13q=−2×(−15)=30x2+13x+30=(x+10)(x+3)so the roots are −3,−10

Two candidates attempt to solve a quadratic of the form $x^{2} + p x + q = 0$ P was taken as 17 in place of 13, its roots were found to be -2 and -15. Find the roots of the original equation.

$x^{2} + p x + q = 0$
$p = 13$
$q = - 2 \times (- 15) = 30$
$x^{2} + 13 x + 30 = (x + 10) (x + 3)$
so the roots are $- 3, - 10$