Question

If a and b are then Zeroes of the 5x^2+2x+3 then a^3 + b^3

Answer 1

$\large{\underline{\bf{\green{Given:-}}}}$

✰ p(x) = 5x² + 2x + 3

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰we need to find the value of α³ + β³

$\huge{\underline{\bf{\red{Solution:-}}}}$

• p(x) = 5x² + 2x + 3

fractorising by middle term splitting :-

$\mapsto \sf\:5x^2+2x+3$

$\mapsto \sf\:5x^2+5x-3x+3$

$\mapsto \sf\:5x(x+1)-3(x+1)$

$\mapsto \sf\:(5x-3)(x+1)$

$\mapsto \sf\:x=\frac{3}{5}\:or\:\:x= -1$

Now,

• Let α = 3/5
• and β = -1

Now finding value of α³ + β³

we know that,

• a ³ + b³ = (a+b)(a² + b² - ab)

Then,

value of α³ + β³ :-

$\mapsto \sf\:[\frac{3}{5}+(-1)][(\frac{3}{5})^2+(-1)^2-(\frac{3}{5}\times(-1)]$

$\mapsto \sf\: (\frac{3 - 5}{5} )( \frac{9}{25} + 1 + \frac{3}{5} ) \\ \\\mapsto \sf\:( \frac{ - 2}{5} )( \frac{9 + 25 + 15}{25} ) \\ \\ \mapsto \sf\: \frac{ - 2}{5} ( \frac{49}{25}) \\ \\ \mapsto \sf\: \frac{ - 98}{125}$

Hence ,

The Value of α³ + β³ = -98/125

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Answer 2

Question :–

• If a and b are then Zeroes of the polynomial 5x² + 2x + 3 = 0 then a³ + b³ = ?

ANSWER :–

$\\ \to \: \: \: \: \: \: { \boxed { \bold{ {a}^{3} + {b}^{3} = { \frac{82}{125}}} }}$

EXPLANATION :–

GIVEN :–

• A quadratic equation 5x² + 2x + 3 = 0 which have two roots a and b.

TO FIND :–

Value of a³ + b³ = ?

SOLUTION :–

▪︎ If a quadratic equation ax² + bx + c = 0 then –

$\\ \: \: \: . \: \: { \bold{Sum \: \: of \: \: roots \: \: = - \dfrac{b}{a} }}$

$\\ \: \: \: . \: \: { \bold{Product \: \: of \: \: roots \: \: = \dfrac{c}{a} }}$

▪︎ Here –

$\\ \: \: \: \: . \: \: { \bold{a =5 }}$

$\\ \: \: \: \: . \: \: { \bold{b =2 }}$

$\\ \: \: \: \: . \: \: { \bold{c =3 }}$

• So that –

$\\ \: \: \: . \: \: { \bold{Sum \: \: of \: \: roots \: \: = a + b = - \dfrac{2}{5} }}$

$\\ \: \: \: . \: \: { \bold{Product \: \: of \: \: roots \: \: = a.b = \dfrac{3}{5} }}$

▪︎ Now let's find (a³ + b³) –

• We know that –

$\\ \implies \: \: \: \: \: { \bold{ {(a + b)}^{3} = {a}^{ 3} + {b}^{3} + 3ab(a + b)}}$

• So that –

$\\ \implies \: \: \: \: \: { \bold{ {a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b)}}$

• Now put the values –

$\\ \implies \: \: \: \: \: { \bold{ {a}^{3} + {b}^{3} = {( - \frac{2}{5} )}^{3} - 3( \frac{3}{5} )( - \frac{2}{5} )}}$

$\\ \implies \: \: \: \: \: { \bold{ {a}^{3} + {b}^{3} = {- \frac{8}{125}} + \frac{18}{25} }}$

$\\ \implies \: \: \: \: \: { \boxed { \bold{ {a}^{3} + {b}^{3} = { \frac{82}{125}}} }}$