Question

0.14 Verify that the indicated values are the zeroes of the quadratic polynomials
(1) x2+ root 2x -4, root 2, -2root 2
(1)x2+2root2 x -6,-3root 2,root 2
Also find the sum and product of the zeroes in each case.​

Answer 1

Given :

The quadratic polynomials are :

• $\sf {x}^{2} + \sqrt{2} x - 4$
• $\sf {x}^{2} + 2 \sqrt{2} x - 6$

Task :

To verify

• √2 , -2√2 are the roots of the polynomial x² + √2x - 4 and
• -3√2 , √2 are the roots of the polynomial x² + 2√2x - 6
• and also to verify the relationship between the coefficients .

Solutions :

$\sf {x}^{2} + \sqrt{2} x - 4 \\ = \sf{x}^{2} - \sqrt{2} x + 2 \sqrt{2} x - 4 \\ = \sf x(x - \sqrt{2} ) + 2 \sqrt{2} x(x - \sqrt{2} ) \\ \sf= (x - \sqrt{2} )(x + 2 \sqrt{2} )$

Thus , the roots are :

x - √2 = 0 and x + 2√2 = 0

→ x = √2 and → x = -2√2

The relationship are :

$\sf{sum \: of \: \: the zeroes = \dfrac{ -coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }$

$\sf \implies \sqrt{2} + ( - 2 \sqrt{2} ) = \dfrac{ - \sqrt{2}}{1} \\ \sf\implies - \sqrt{2} = - \sqrt{2}$

and again ,

$\sf{product \: of \: zeroes = \dfrac{constant}{coefficient \: of \: {x}^{2} } }$

$\sf \implies( \sqrt{2} )( - 2 \sqrt{2} ) = \dfrac{ - 4}{1} \\ \sf \sf\implies - 4 = - 4$

___________________________

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

Thus , the roots are :

x - √2 = 0 and x + 3√2 = 0

→ x = √2 and → x = -3√2

Now the relations of coefficients

$\sf{sum \: of \: \: the zeroes = \dfrac{ -coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }$

$\sf \sqrt{2} - 3 \sqrt{2} = - \dfrac{2 \sqrt{3} }{1} \\ \sf\implies - 2 \sqrt{2} = - 2 \sqrt{2}$

and

$\sf{product \: of \: zeroes = \dfrac{constant}{coefficient \: of \: {x}^{2} } }$

$\sf\sf( \sqrt{2} )( - 3 \sqrt{2} ) = \sf\dfrac{ - 6}{1} \\ \implies - 6 = - 6$

$\bf{Hence \: \: verified }$

Answer 2

Answer:

$\underline{ \bf{\dag}\:\:\large{\textit{Question 1 :}}}$

$:\implies\sf {x}^{2} + \sqrt{2} x - 4 \\\\\\:\implies\sf {x}^{2} - \sqrt{2} x + 2 \sqrt{2} x - 4 \\\\\\:\implies\sf x(x - \sqrt{2} ) + 2 \sqrt{2} x(x - \sqrt{2} ) \\\\\\:\implies\sf(x - \sqrt{2} )(x + 2\sqrt{2} )\\\\\\:\implies\sf x=\sqrt{2}\quad or\quad x=-\:2\sqrt{2}$

$\rule{100}{1}$

$\underline{\bigstar\:\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{2} + (-\:2\sqrt{2}) = \dfrac{-\sqrt{2}}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt -\:\sqrt{2}=-\:\sqrt{2}}}}\\\\\\{\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{2} \times (-\:2\sqrt{2}) = \dfrac{-4}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt -\:4 =-\:4}}}$

$\rule{200}{2}$

$\underline{ \bf{\dag}\:\:\large{\textit{Question 2 :}}}$

$:\implies\sf {x}^{2} + 2 \sqrt{2} x - 6 \\\\\\:\implies\sf x + 3 \sqrt{2} x- \sqrt{2} x - 6 \\\\\\:\implies\sf x(x + 3 \sqrt{2} ) - \sqrt{2} (x + 3 \sqrt{2} )\\\\\\:\implies\sf (x - \sqrt{2} )(x + 3 \sqrt{2} )\\\\\\:\implies\sf x=\sqrt{2}\quad or\quad x=-\:3\sqrt{2}$

$\rule{100}{1}$

$\underline{\bigstar\:\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{2} + (-\:3\sqrt{2}) = \dfrac{-2\sqrt{2}}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt -\:2\sqrt{2}=-\:2\sqrt{2}}}}\\\\\\{\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \sqrt{2} \times (-\:3\sqrt{2}) = \dfrac{-6}{1}\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt -\:6 =-\:6}}}$