Question

Find the zeros of each of the following quadratic polynomials and verify the relationship the zeros and their coefficients: p(x)=x²- 2×root2x -6

Answer 1

Given

The quadratic polynomial is :

• x² - 2√2x - 6

Task

• To find the zeroes of the polynomial
• To verify the relationship between the zeroes and the coefficients of the polynomial

Solution

$\rm {x}^{2} - 2 \sqrt{2} x - 6 \\ \\ = \rm{x}^{2} + \sqrt{2} x - 3 \sqrt{2} x - 6 \\ \\ = \rm x(x + \sqrt{2} ) - 3 \sqrt{2} (x + \sqrt{2} ) \\ \\ = \rm(x - 3 \sqrt{2} )(x + \sqrt{2} )$

Now the zeroes of the polynomial are :

$\rm x - 3\sqrt{2} = 0 \: \: and \: \: x + \sqrt{2}=0 \\\\ \rm\implies x = 3\sqrt{2} \: \: and \: \implies x = -\sqrt{2}$

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Now verification of the relationships :

$\rm{sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: x {}^{2} } } \\ \\ \implies3 \sqrt{2} + ( - \sqrt{2} ) = - (\frac{ - 2 \sqrt{2} }{1} ) \\ \\ \implies2 \sqrt{2} = 2 \sqrt{2}$

Again we have

$\rm product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} } \\ \\ \implies(3 \sqrt{2} )( - \sqrt{2} ) = \frac{ - 6}{1} \\ \\ \implies - 6 = - 6$

Hence Verified