When b=17,it is found that the roots of the e | vedclass

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(A) Given the quadratic equation $x^2+bx+c=0$ with $b=17$. Since the roots are $-2$ and $-15$,the sum of roots is $-2 + (-15) = -17$. From the equatio

Vedclass · Accepted Answer

$7$

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(A) Given the quadratic equation $x^2+bx+c=0$ with $b=17$. Since the roots are $-2$ and $-15$,the sum of roots is $-2 + (-15) = -17$. From the equation $x^2+bx+c=0$,the sum of roots is $-b$. Thus,$-b = -17$,which matches $b=17$. The product of roots is $c = (-2) 	imes (-15) = 30$. Now,consider the equation with $b=13$ and $c=30$: $x^2+13x+30=0$. Factoring the quadratic: $x^2+10x+3x+30=0 \implies (x+10)(x+3)=0$. The roots are $\alpha = -10$ and $\beta = -3$. Then,$|\alpha-\beta| = |-10 - (-3)| = |-10+3| = |-7| = 7$. Therefore,the correct option is $A$.

When $b=17$,it is found that the roots of the equation $x^2+bx+c=0$ are $-2$ and $-15$. If $\alpha, \beta$ are the roots of the same equation when $b=13$,then $|\alpha-\beta|=$

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