Question

If a and b are the roots of quadratic equation 3m²+ 2m – 4 = 0, Then value of a²+ b² is​

Answer 1

$\Large {\underline { \sf {Answer :}}}$

$\sf{ \dfrac{28}{9} }$

$\Large {\underline { \sf {Explication \; of \; steps :}}}$

As per the provided information in the given question, we have :

• Given quadratic equation = 3m²+ 2m + (– 4) = 0
• Here, $\alpha$ and $\beta$ are the roots of it.

We are asked to calculate the value of $\sf { ( \alpha^2 + \beta^2) }$ .

We know that, if $\alpha$and $\beta$ are roots of the quadratic equation$\sf{ ax^2+ bx + c = 0}$. Then,

$\longrightarrow \underline{\boxed{ \sf { \alpha + \beta = \dfrac{-b}{a} }}}$

and,

$\longrightarrow\underline{\boxed{ \sf { \alpha \times \beta = \dfrac{c}{a} }}}$

Here,

• a = 3
• b = 2
• c = -4

Substituting the values in both formulae :

$\longrightarrow \sf { \alpha + \beta = \dfrac{-b}{a} }$

$\longrightarrow \sf { \alpha + \beta = \dfrac{-2}{3} }$

Let it be the equation (1).

$\longrightarrow \sf { \alpha \times \beta = \dfrac{c}{a} }$

$\longrightarrow \sf { \alpha \times \beta = \dfrac{-4}{3} }$

Let it be the equation (2).

Now, we have,

$\longrightarrow \sf { \alpha + \beta = \dfrac{-2}{3} }$

On squaring both sides,

$\longrightarrow \sf { {(\alpha + \beta)}^{2} = {\Bigg \lgroup \dfrac{-2}{3} \Bigg \rgroup}^{2} }$

$\longrightarrow \sf {( \alpha)^2 + (\beta)^2 + 2( \alpha \times \beta ) = \dfrac{4}{9} }$

Substituting the value of $(\alpha \times \beta )$ from the equation (2).

$\longrightarrow \sf {( \alpha)^2 + (\beta)^2 + 2\Bigg ( \dfrac{-4}{3} \Bigg ) = \dfrac{4}{9} }$

$\longrightarrow \sf {\alpha^2 + \beta^2 + \Bigg ( \dfrac{-8}{3} \Bigg ) = \dfrac{4}{9} }$

$\longrightarrow \sf {\alpha^2 + \beta^2 - \dfrac{8}{3} = \dfrac{4}{9} }$

Transposing -8/3 from L.H.S to R.H.S.

$\longrightarrow \sf {\alpha^2 + \beta^2 = \dfrac{4}{9} + \dfrac{8}{3} }$

$\longrightarrow \sf {\alpha^2 + \beta^2 = \dfrac{4+24}{9} }$

$\longrightarrow \underline{\boxed{ \sf {\alpha^2 + \beta^2 = \dfrac{28}{9}} }} \; \bigstar$

$\therefore\; \sf { {(\alpha + \beta)}^{2} }$ is 28/9.

$\Large {\underline { \sf {More \; Information :}}}$

• Quadratic equation is an equation of second degree or degree 2, so it has two roots.

• If $\alpha$ and $\beta$ satisfy the equation $\sf{ ax^2+ bx + c = 0}$ then If $\alpha$ and $\beta$ are called roots of the equation.

• Roots of quadratic equation are also called zeroes of the quadratic equation.

• If $\alpha$ and $\beta$ are roots of the quadratic equation$\sf{ ax^2+ bx + c = 0}$. Then,

$\longrightarrow \underline{\boxed{ \sf { \alpha + \beta = \dfrac{-b}{a} }}}$

and,

$\longrightarrow\underline{\boxed{ \sf { \alpha \times \beta = \dfrac{c}{a} }}}$