Question

Find a quadratic polynomial, the sum of whose
zeroes is 7 and their product is 12. Hence find the
zeroes of the polynomial

Answer 1

Step-by-step explanation:

let a and b be the zeros of polynomial

according to question,

a+b=7

ab=12

then the required polynomial is,

${x}^{2} - (a + b)x + ab = 0$

${x}^{2} - 7x + 10 = 0$

Answer 2

Given:

Sum of zeroes = 7

Product of zeroes = 12

To find:

• Quadratic polynomial
• Zeroes of the polynomial

Solution:

Finding quadratic polynomial

${x}^{2} - (sum \: of \: zeroes) + (product \: of \: zeroes)$

Given that,

Sum of zeroes = 7

Product of zeroes = 12

Substituting values,

$\implies \: {x}^{2} - (7)x + (12) \:$

Hence, the required polynomial is x² - 7x + 12.

Finding zeroes of the polynomial

We have,

${x}^{2} - 7x + 12$

By splitting the middle term

${x}^{2} - 4x - 3x + 12 = 0$

$x(x - 4) - 3(x - 4) = 0$

$(x - 3)(x - 4) = 0 \\ \\ x = 3 \: and \: x = 4$

Hence, zeroes of the polynomial are 3 and 4.

Verification:

Sum of zeroes = 7 (given)

Also, sum of zeroes

=> 3+4 = 7

And,

Product of zeroes = 12 (given)

Also, product equals

=> 3 × 4 = 12

Hence verified.

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