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Standard XII

Mathematics

Roots of a Quadratic Equation

The coefficie...

Question

The coefficient of $x$ in the equation $x^{2} + p x + q = 0$ was wrongly written as $17$ in place of $13$ and the roots thus found was $- 2$ and $- 15$ .
Then the roots of the correct equation are

- 3, 10

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- 3, - 10

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3, - 10

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None of the above.

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Solution

The correct option is C $- 3, - 10$
$x^{2} + p x + q = 0 p = 13 q = (- 2) \times (- 15) = 30 x^{2} + 13 x + 30 = 0 (x + 3) (x + 10) = 0 x = - 10, - 3$

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Similar questions

Q. The coefficient of

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x^{2} + p x + q = 0

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and the roots thus found to be

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The coefficient of x in the equation $x^{2}$ + p $x$ + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

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Q. The coefficient of x in the quadratic equation

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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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The coefficient of x in the equation x2+px+q=0 was wrongly written as 17 in place of 13 and the roots thus found was −2 and −15.Then the roots of the correct equation are

The correct option is C −3,−10x2+px+q=0p=13q=(−2)×(−15)=30x2+13x+30=0(x+3)(x+10)=0x=−10,−3

The coefficient of $x$ in the equation $x^{2} + p x + q = 0$ was wrongly written as $17$ in place of $13$ and the roots thus found was $- 2$ and $- 15$ .
Then the roots of the correct equation are

The correct option is C $- 3, - 10$
$x^{2} + p x + q = 0 p = 13 q = (- 2) \times (- 15) = 30 x^{2} + 13 x + 30 = 0 (x + 3) (x + 10) = 0 x = - 10, - 3$