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Question Number 210362 by MATHEMATICSAM last updated on 10/Sep/24

$$\mathrm{If}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{quadratic}\mathrm{equation}$$$$(a-b+c)x^2+(c-b-a)x+2(b-c)=0$$$$\mathrm{are}\mathrm{real}\mathrm{and}\mathrm{equal}\mathrm{then}\mathrm{find}\frac{a}{b-c}.$$

Commented by som(math1967) last updated on 08/Aug/24

$$notqudratic,itshouldbe(a-b+c)x^2$$

Commented by MATHEMATICSAM last updated on 10/Sep/24

$$\mathrm{oh}\mathrm{yes}$$

Answered by som(math1967) last updated on 11/Sep/24

$$let(a-b+c)=p,(c-b-a)=q$$$$2(b-c)=r$$$$\therefore p+q+r=a-b+c+c-b-a+2b-2c$$$$=0$$$$equation\Rightarrow {px}^2+qx+r=0$$$$rootsarerealandequal$$$$\therefore q^2=4pr$$$$\Rightarrow {(p+r)}^2=4pr[\because p+q+r=0]$$$$\Rightarrow {(p-r)}^2=0$$$$\Rightarrow p-r=0\Rightarrow p=r$$$$\Rightarrow a-b+c=2(b-c)$$$$\Rightarrow a+c+2c=2b+b$$$$\Rightarrow a=3(b-c)$$$$\therefore \frac{a}{b-c}=3$$

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