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प्रश्न
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.
4u2 + 8u
योग
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उत्तर
4u2 + 8u = 4u(u + 2)
= 4[u - 0][u - (-2)]
For p(u) = 0, we have
Either 4u = 0
u = -2
∴ The zeroes of 4u2 + 8u are 0 and -2.
Relationship between the zeroes and the coefficients of the polynomial
Sum of the zeroes = `-("Coefficient of " u)/("Coefficient of " u^2)`
= `0 + (-2) =(-(8))/4`
= -2 = -2
Also product of the zeroes = `"Constant term"/("Coefficient of " u^2)`
= `0 xx (-2) = 0/4`
= 0 = 0
Thus, the relationship between the zeroes and the coefficients in the polynomial 4u2 + 8u is verified.
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क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Polynomials - EXERCISE 2.2 [पृष्ठ ३३]
