Question

find the roots of the following equations​

Answer 1

Answer:

L.H.S

= √2x² + 7x + 5√2.

= √2x² + 2x + 5x + 5√2.

= (√2x² + 2x) + ( 5x + 5√2).

=. ( √2x² + √2 × √2 × x) + ( 5x + 5√2).

=. √2x( x + √2 ) + 5( x + √2 ).

= ( x + √2 ) ( √2x + 5 )

1. x + √2 = 0

x = - √2

2. √2x + 5 = 0

√2x = - 5

x = -5/√2.

Step-by-step explanation:

solve by splitting the middle term.

Answer 2

Answer:

• $\rm -\sqrt{2} \: and \:\dfrac{ - 5\sqrt{2}}{ {2} }$

Explanation:

Given equation,

$= \boldsymbol { \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0 }$

On comparing with the general form of a quadratic equation $\boldsymbol {ax^2 + bx + c = 0}$

We get,

• $\boldsymbol {a = \sqrt{2}}$
• $\boldsymbol {b = 7}$
• $\boldsymbol {c = 5\sqrt{2}}$

Using quadratic formula,

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

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

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

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

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

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

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

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

On rationalising the denominator,

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

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