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In solving a problem that reduces to a quadratic equation, one student makes a mistake in the constant term and obtains 8 and 2 for roots. Another student makes a mistake only in the coefficient of first-degree term and finds $-9$ and $-1$ for roots. The correct equation is
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The correct answer is:
$x^{2}-10 x+9=0$
Letcorrect equation is $a x^{2}+b x+c=0$ According to first student, equation is $a x^{2}+b x+c_{1}=0$ and roots are 8 and
$8+2=\frac{-b}{a} \Rightarrow \frac{b}{a}=-10$
Quadratic equation according to second student $a x^{2}+b_{1} x+c=0$ and roots are $-9$ and $-1$
$(-9) \times(-1)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{c}}{\mathrm{a}}=9$
Putting value in original equation
$x^{2}+\frac{b}{a} x+\frac{c}{a}=0$
$x^{2}-10 x+9=0$
$8+2=\frac{-b}{a} \Rightarrow \frac{b}{a}=-10$
Quadratic equation according to second student $a x^{2}+b_{1} x+c=0$ and roots are $-9$ and $-1$
$(-9) \times(-1)=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{c}}{\mathrm{a}}=9$
Putting value in original equation
$x^{2}+\frac{b}{a} x+\frac{c}{a}=0$
$x^{2}-10 x+9=0$
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