Question

Answer both of the questions

Answer 1

ANSWER.


GIVEN.

QUADRATIC EQUATIONS

PART 1


FIND ROOT OF. QUADRATIC EQUATION

${x}^{2 } + 3ax + 2 {a}^{2} = 0$


${x}^{2} + 2ax + ax + 2 {a}^{2} = 0$


$x(x + 2a) + a(x + 2a) = 0$


ROOTS

$(x + 2a) \: \: \: (x + a) \: \:$

$\alpha = - 2a \: \: \: \: and \: \beta = - a$


NOW. it is. given

${ \alpha }^{2} + { \beta }^{2} = 5$


=>.
${( - 2a)}^{2} + { (- a)}^{2} = 5$


$4 {a}^{2} \: + {a}^{2} = 5$


$5 {a}^{2} = 5$
${a}^{2} = \frac{5}{5}$


${a}^{2} = 1$
$a = \frac{ + }{ - } 1$
a=1, a =-1

PART 2

DISCRIMINATE =
${b}^{2} - 4ac$
b=3a

b= (3×1)= 3. , b=( 3×-1)=-3

so it value for given eq

and a=1 and c=2

$= > {3}^{2} -( 4 \times 1 \times 2)$

$= 9 - 8 = 1$

1. is positive



and second. DISCRIMINATE
$= {b}^{2} -( 4 \times a \times c)$


=
${( - 3)}^{2} -( 4 \times 1 \times 2)$

$= 9 - 8 = 1$


so DISCRIMINATE IS POSITIVE

$= > d > 0$


Hence


ANS

(1) a= 1,-1

(2)D>0

Answer 2

Answer:

79) 3) +-1

80) 1) D>0

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