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Find the quadratic polynomial, sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
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Let ЁЭЫ╝ and ЁЭЫ╜ be the zeroes of the required polynomial f(x).
Then (ЁЭЫ╝ + ЁЭЫ╜) = 8 and ЁЭЫ╝ЁЭЫ╜ = 12
∴ `f(x)=x^2-(∝+β)x+∝β `
`⇒ f(x)=x^2-8x+12`
Hence, required polynomial `f(x)=x^2-8x+12`
`∴ f(x)=0 ⇒ x^2-8x+12=0`
`⇒ x^2-(6x+2x)+12=0`
`⇒ x^2-6x-2x+12=0`
`⇒x(x-6)-2(x-6)=0`
`⇒ (x-2) (x-6)=0`
`⇒ (x-2)=0 or (x-6)=0`
`⇒x=2 or x=6`
So, the zeroes of f(x) are 2 and 6.
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Case Study -1
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