Question

al find zeros of quadratic equation and verify the relastionship between them. √2n²+7x+ 5√2​

Answer 1

Answer:

${ \sf{ \underline{ \: Zeros \: of \:the \: polynomial \: are \: \dfrac{ - 2}{ \sqrt{2} } \: and \dfrac{ - 5}{ \sqrt{2} } }}}$

Explanation:

Correct polynomial,

$\rm \implies \: \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2}$

On comparing with the general form of a quadratic equation ax² + bx + c

We get,

• a = √2
• b = 7
• c = 5√2

By quadratic formula,

$\rm \implies \: x = \dfrac{ - b \pm \sqrt{ {(b)}^{2} - 4ac } }{2a}$

$\rm \implies \: x = \dfrac{ - 7 \pm \sqrt{ {(7)}^{2} - 4( \sqrt{2} )(5 \sqrt{2}) } }{2( \sqrt{2} )}$

$\rm \implies \: x = \dfrac{ - 7 \pm \sqrt{49 - 40 } }{2 \sqrt{2} }$

$\rm \implies \: x = \dfrac{ - 7 \pm \sqrt{9 } }{2 \sqrt{2} }$

$\rm \implies \: x = \dfrac{ - 7 \pm 3 }{2 \sqrt{2} }$

$\rm \implies \: x = \dfrac{ - 7 + 3 }{2 \sqrt{2} } \: and \: \dfrac{ - 7 - 3}{2 \sqrt{2} }$

$\rm \implies \: x = \dfrac{ - 4}{2 \sqrt{2} } \: and \: \dfrac{ -10}{2 \sqrt{2} }$

$\rm \implies \: x = \dfrac{ - 2}{ \sqrt{2} } \: and \: \dfrac{ -5}{ \sqrt{2} }$

${ \sf{ \underline{ \therefore \: Zeros \: of \:the \: polynomial \: are \: \dfrac{ - 2}{ \sqrt{2} } \: and \dfrac{ - 5}{ \sqrt{2} } }}}$

Verification:

We know,

$\rm \bullet \: sum \: of \: zeros = \dfrac{ - b}{a}$

$\rm \implies \: \dfrac{ - 2}{ \sqrt{2} } + \dfrac{ - 5}{ \sqrt{2} } = \dfrac{ - 7}{ \sqrt{2} }$

$\rm \implies \: \dfrac{ - 2 + ( - 5)}{ \sqrt{2} } = \dfrac{ - 7}{ \sqrt{2} }$

$\rm \implies \: \dfrac{ - 2 - 5}{ \sqrt{2} } = \dfrac{ - 7}{ \sqrt{2} }$

$\rm \implies \: \dfrac{ - 7}{ \sqrt{2} } = \dfrac{ - 7}{ \sqrt{2} }$

And also,

$\rm \bullet \: product \: of \: zeros \: = \dfrac{c}{a}$

$\rm \implies \: \dfrac{ - 2}{ \sqrt{2} } \ast \dfrac{ - 5}{ \sqrt{2} } = \dfrac{5 \sqrt{2} }{ \sqrt{2} }$

$\rm \implies \: \dfrac{ 10}{ {2} } = 5$

$\rm \implies \: 5 = 5$

Hence, verified!!