Question

Find the zeroes of the polynomial x2 -7X+12and verify the relationship between the zeroes and the coefficients.​

Answer 1

Given :

• Polynomial - $\sf\:x^{2}-7x+12$

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To find :

• Zeroes of the polynomial.
• Verify the relationship between the zeroes and the coefficient.

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Formula to be used :

For any quadratic equation ax² + bx + c = 0 having roots as α and β -

Sum of the roots -

$\bf\color{purple}{\alpha + \beta = \frac{-b}{a}}$

Product of the roots -

$\bf\color{purple}{\alpha \times \beta = \frac{c}{a}}$

We would use this formulae to verify the relationship between the zeroes.‎

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Solution :

Calculating the zeroes if the polynomial by middle term breaking -

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

$\sf\implies\:x^{2}-(4+3)x+12=0$

$\sf\implies\:x^{2}-4x-3x+12=0$

$\sf\implies\:x(x-4)-3(x-4)=0$

$\sf\implies\:(x-4)(x-3)=0$

Either :

$\sf\:x-4=0$ $\small{\underline{\boxed{\mathrm\purple{\dashrightarrow\:x~=~4}}}}$

Or :

$\sf\:x-3=0$ $\small{\underline{\boxed{\mathrm\purple{\dashrightarrow\:x~=~3}}}}$

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$\sf\therefore\:Two~zeroes~of~the~polynomial~are~4~and~3$

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Verifying the relationship b/w the zeroes and the coefficient -

In the given polynomial - $\sf\:x^{2}-7x+12$

• a = $\bf\:1$ , b = $\bf\:(-7)$ and c = $\bf\:12$

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$\bf\maltese$ Sum of the zeroes -

$\sf\twoheadrightarrow\:\alpha+\beta~=~\frac{-b}{a}$

$\sf\twoheadrightarrow\:4+3~=~\frac{-(-7)}{1}$

$\sf\twoheadrightarrow\:7~=~\frac{7}{1}$

$\small{\underline{\boxed{\mathrm{\twoheadrightarrow\:7~=~7}}}}$

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$\bf\maltese$ Product of the zeroes -

$\sf\twoheadrightarrow\:\alpha \times \beta~=~\frac{c}{a}$

$\sf\twoheadrightarrow\:4 \times 3~=~\frac{12}{1}$

$\small{\underline{\boxed{\mathrm{\twoheadrightarrow\:12~=~12}}}}$

Hence Verified !~‎

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