Question

4 find the zero
of the quadrate polynomial x2+7x+12 and Venfy the relation between the zero
and its co-efficient.

Answer 1

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☆ Correct Question:-

QUE:- Find the zeroes

of the quadratic polynomial x²+7x+12 and Verify the relation between the zero

and its co-efficient.

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☆ ANSWER:-

▪︎ Given quadratic polynomial:-

$\tt\leadsto {x}^{2} + 7x + 12.$

▪︎ To Find:-

• The zeroes of the given polynomial.

• And, to verify the relationship between the zeroes and coefficients of a quadratic polynomial.

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▪︎ Concepts Used:-

$\tt\leadsto \small\: sum \: of \: zeroes = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } = ( \frac{ - b}{a}) \\ \\ \tt\leadsto \small\: product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } = (\frac{c}{a} )$

▪︎ So let us first find the zeroes:-

• Since we are finding the zeroes of a polynomial so for this it should be equal to 0.

$\sf \mapsto {x}^{2} + 7x + 12 = 0$

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• And, by factorization method,

• We have to first find the product of coefficient of x² and constant term,

= 1 × 12 = 12.

• Now lets factorize 12.

=》 12 = 2×2×3.

• Also,

=》 (2×2)+3 = 4+3 = 7 = Middle term of the given polynomial.

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▪︎ Therefore lets split the middle term now,

$\sf \mapsto {x}^{2} + (4 + 3)x + 12 = 0 \\ \\ \sf \mapsto {x}^{2} + 4x + 3x + 12 = 0 \\ \\ \sf \mapsto x(x + 4) + 3(x + 4) = 0 \\ \\ \sf \mapsto(x + 3)(x + 4) = 0 \\ \\ \sf \large\red{\fbox{\mapsto \: either \: \: x = - 3 \: \: or \: \: x = - 4}}$

▪︎ Therefore the required value of x are -3 and -4.

• So the zeroes of the polynomial are -3 and -4.

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☆ VERIFICATION:-

▪︎ Here,

• Coefficient of x² = 1.
• Coefficient of x = 7.
• Constant term = 12.

▪︎ Therefore,

• Sum of zeroes should be equal to -7 and product of zeroes should be equal to 12.

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▪︎ So lets verify:-

◇ Sum Of zeroes,

$\sf = ( - 4) + ( - 3) \\ \\ = - 4 - 3 \\ \\ = - 7 = \frac{ - b}{a}$

◇ Product of zeroes,

$\sf = ( - 4) \times ( - 3) \\ \\ = 12 = \frac{c}{a}$

☆ Hence Verified.

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