Question

Answer the question in the attachment

No spam plz ​

Answer 1

Answer:

So easy it is girl

Step-by-step explanation:

You can see

any promblem contact me

Answer 2

$\qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{Given \; that :}}}}}}}$

★ Quadratic Equation : ax² + bx + c = 0

$\qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{To \; Find :}}}}}}}$

★ The value of  the expression 1/α² + 1/β²

$\qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{Assumptions :}}}}}}}$

★ Let the roots of the equation be considered as alpha ( α ) and beta ( β )

$\qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{Basic \; Knowledge :}}}}}}}$

★ The product of α and β is equal to c / a

★ The sum of α and β is equal to - b / a

$\qquad \qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{Required \; Solution:}}}}}}}$

Now here we have been given the general form of an quadratic equation ( the basic structure of the equation which is ax² + bx + c and said to find out the value of :

$\bigstar \; {\underline {\boxed{\tt {\dfrac{1}{\alpha ^2} + \dfrac{1}{\beta^2} }} }}$

Now as we know the basic formula which state the relation between the roots and the constant terms in the equation

•   αβ = c / a
•   α + β = - b / a

So, let's use the formulas converting the required value into such suitable forms where the following formula can be embed and find the value required

$: \implies \bf \dfrac{1}{\alpha^2} + \dfrac{1}{\beta ^2}$

Now let's take L.C.M and add up the fractions :

$: \implies \sf \dfrac{1 ( \beta^2 ) + 1 (\alpha^2 )}{\alpha^2 \beta^2} \\ \\ \\ : \implies \sf \dfrac{\alpha^2 + \beta^2}{\alpha^2\beta^2} \\ \\ \\ : \implies \sf \dfrac{\alpha^2 + \beta^2}{(\alpha\beta )^2}$

Now, let's use the identity of ( a + b ) ² which is that ,

$\qquad \bigstar \; {\underline {\boxed{\bf {( a + b ) ^2 = a^2 + b ^2 + 2ab} }} }$

Let's simplify the numerator part of the fraction assuming α as a and β and b and split it into suitable forms to apply the formulas we have

$: \implies \sf \dfrac{\alpha^2 + \beta^2}{(\alpha\beta )^2} \\ \\ \\ : \implies \sf \dfrac{ (\alpha + \beta )^2 = \alpha^2 + \beta^2 + 2\alpha \beta}{(\alpha\beta )^2} \\ \\ \\: \implies \sf \dfrac{ (\alpha + \beta )^2 - 2 \alpha\beta= \alpha^2 + \beta^2 }{(\alpha\beta )^2}$

Now it can be rewritten in the form such as :

$\bigstar \; {\underline {\boxed{\tt {\dfrac{(\alpha + \beta)^2 - 2 \alpha \beta}{(\alpha \beta ) ^2} }} }}$

Applying the formulas which state the relation of thee roots and the constants we get,

$: \implies \tt \dfrac{ \bigg( \dfrac{-b}{a} \bigg)^2 - 2 \bigg[ \dfrac{c}{a} \bigg]}{\bigg(\dfrac{c}{a}\bigg )^2} \\ \\ \\ : \implies \tt \dfrac{ \dfrac{b^2}{a^2} - \dfrac{2c}{a} }{\dfrac{c^2}{a^2}} \\ \\ \\ : \implies \tt \dfrac{ \dfrac{b^2- 2ac}{\cancel{a^2}} }{\dfrac{c^2}{\cancel{a^2}}} \\ \\ \\ : \implies \tt \dfrac{ b^2 - 2ac }{c^2}$

Therefore:

• ${\underline {\boxed{\tt {\dfrac{1}{\alpha ^2} + \dfrac{1}{\beta^2} = \dfrac{b^2 - 2ac}{c^2} }} }\bigstar}$

$\qquad \qquad {\maltese \; {\underline{\purple{\underline{\pmb{\bf{More \; To \; Know :}}}}}}}$

$\begin{gathered}\boxed{\begin{array}{c} \\ \tiny\bf{\dag}\:\underline{\frak{\rm{S}\frak{ome\:important\:algebric\:identities\:::}}} \\\\ \green{\bigstar}\:\rm \red{ (A+B)^{2} = A^{2} + 2AB + B^{2}} \\\\ \red{\bigstar}\rm\: \green{(A-B)^{2} = A^{2} - 2AB + B^{2}} \\\\ \orange{\bigstar}\rm\: \blue{A^{2} - B^{2} = (A+B)(A-B)}\\\\ \blue{\bigstar}\rm\: \orange{(A+B)^{2} = (A-B)^{2} + 4AB}\\\\ \pink{\bigstar}\rm\: \purple{(A-B)^{2} = (A+B)^{2} - 4AB}\\\\ \purple{\bigstar} \rm\: \pink{(A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}}\\\\ \gray{\bigstar}\rm\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\ \bigstar\rm\: \gray{A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})} \\\\ \end{array}}\end{gathered}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━