An algebra problem by karthik dontula

Algebra Level 2

In a x 2 + b x + c = 0 a{ x }^{ 2 }+bx+c=0 , if a + c = b a+c= b then the roots of the equation are

1 and c/a -1 and -c/a -1 and c/a 1 and -c/a

Karthik Dontula by karthik dontula , 19, India

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1 solution

By Vieta's Formula.
• α + β = − b a \alpha+\beta=\frac{-b}{a} ...... ( 1 ) (1)
• α × β = c a \alpha×\beta=\frac{c}{a} ........ ( 2 ) (2)
Here, α \alpha and β \beta are roots.
Since, a + c = b a+c=b or − b = − a − c -b=-a-c or c = b − a c=b-a
From ( 1 ) (1) ,
α + β = − a − c a \alpha+\beta=\frac{-a-c}{a} = 1 − c a 1-\frac{c}{a} [ By comparing α = − 1 \alpha=-1 and β = − c a \beta=\frac{-c}{a} ]

Now checking.
From ( 2 ) 2) ,
α × β \alpha×\beta = − 1 × − c a -1×\frac{-c}{a} = c a \frac{c}{a} .
Therefore, both vieta's equation satisfies.
Therefore, Roots are − 1 -1 and − c a \frac{-c}{a}

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