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Nature of Roots

The coefficie...

Question

The coefficient of $x$ in the equation $a x^{2} + b x + c = 0$ was wrongly taken as $17$ instead of $13$ & its roots were $- 2, - 15$ . Actual roots are _____.

- 2, 15

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$- 3, - 10$

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$- 4, - 9$

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$- 5, - 6$

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Solution

The correct option is B
$- 3, - 10$

Product of the roots $= 30$
Therefore equation becomes
$x^{2} + 13 x + 30$
$= (x + 10) (x + 3)$
Therefore, the roots are $x = - 10, - 3$

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The coefficient of $x$ in the equation $a x^{2} + b x + c = 0$ was wrongly taken as $17$ instead of $13$ & its roots were $- 2, - 15$ . Actual roots are _____.

Q. The coefficient of x in the equation

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was wrongly taken as 17 instead of 30 and its roots came out to be -2, -15. The actual roots were:

Q. The coefficient of

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in the equation

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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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The coefficient of x in the equation ax2+bx+c=0 was wrongly taken as 17 instead of 13 & its roots were −2,−15. Actual roots are _____.

The correct option is B −3,−10 Product of the roots =30 Therefore equation becomes x2+13x+30 =(x+10)(x+3) Therefore, the roots are x=−10,−3

The coefficient of $x$ in the equation $a x^{2} + b x + c = 0$ was wrongly taken as $17$ instead of $13$ & its roots were $- 2, - 15$ . Actual roots are _____.

The correct option is B
$- 3, - 10$

Product of the roots $= 30$
Therefore equation becomes
$x^{2} + 13 x + 30$
$= (x + 10) (x + 3)$
Therefore, the roots are $x = - 10, - 3$