Question

The roots of the equation x2 + 6x - 4 = 0 are α, β. Find the quadratic equation whose roots are α2 and β2

Answer 1

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

• p(x) = x² + 6x - 4 = 0

$\large\bf\underline \orange{To \: find:-}$

• Find the quadratic equation whose roots are α2 and β2

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

★ Given polynomial = x² + 6x - 4 = 0

• a = 1
• b = 6
• c = -4

• α + β = -b/a

⠀⠀⠀⠀⠀➝α + β = -6/1

⠀⠀⠀⠀⠀➝α + β = -6

• αβ = c/a

⠀⠀⠀⠀⠀➝ -4/1

⠀⠀⠀⠀⠀➝ -4

• α² + β² = (α + β)² - 2αβ

⠀⠀⠀⠀⠀➝ α² + β² = (-6)² - 2× -4

⠀⠀⠀⠀⠀➝ α² + β² = 36 + 8

⠀⠀⠀⠀⠀➝ α² + β² = 44

The quadratic polynomial whose roots are α² and β²

★ Formula for quadratic equation :-

• x² - (α + β)x + αβ

⠀⠀⠀⠀⠀➝ x² - (α² + β²)x + α²β²

⠀⠀⠀⠀⠀➝ x² - (44)x + (-4)²

⠀⠀⠀⠀⠀➝ x² - 44x + 16

So,

the quadratic equation whose roots are α2 and β2 is ★ x² - 44x + 16.

Answer 2

Question :–

• The roots of the equation x² + 6x - 4 = 0 are α, β. Find the quadratic equation whose roots are α² and β².

ANSWER :–

GIVEN :–

• A quadratic equation x² + 6x - 4 = 0 have two roots α, β .

TO FIND :–

• Another quadratic equation whose roots are α² and β².

SOLUTION :–

• A quadratic equation ax² + bx + c = 0 whose roots are α and β , then –

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{sum \: \: of \: \: roots \: \: = \alpha + \beta = - \dfrac{b}{a} }}$

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{product \: \: of \: \: roots \: \: = \alpha . \beta = \dfrac{c}{a} }}$

• Here –

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{a = 1 }}$

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{b = 6 }}$

$\\ \: \: \: \: \: { \huge{.}} \: \: { \bold{c = -4 }}$

• So that –

$\\ \longrightarrow \: \: { \bold{sum \: \: of \: \: roots \: \: = \alpha + \beta = - \dfrac{6}{1} }}$

$\\ \: \: \: \: \: \: { \huge{.}} \: \: { \bold{ \alpha + \beta = - 6 \: \: \: \: \: - - - - eq.(1)}}$

$\\ \longrightarrow \: \: { \bold{product \: \: of \: \: roots \: \: = \alpha . \beta = \dfrac{ - 4}{1} }}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{ \alpha . \beta = - 4 \: \: \: \: \: - - - - eq.(2)}}$

☞ Now Another quadratic equation whose roots are α² and β² –

$\\ \longrightarrow \: \: { \bold{sum \: \: of \: \: roots \: \:of \: \: required \: \: \: ploynomial = { \alpha }^{2} + { \beta }^{2} }}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{sum \: \: of \: \: roots = {( \alpha + \beta ) }^{2} - 2 \alpha \beta \: \: \: [ \: \because \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab ]}}$

$\\ \: \: \: \: \: \: \:{ \huge{.}} \: \: { \bold{sum \: \: of \: \: roots = {( - 6) }^{2} - 2 ( - 4) \: \: \: \: \: \: [ \: using \: \: eq.(1) \: \: and \: \: (2) ]}}$

$\\ \: \: \: \: \: \: \:{ \huge{.}} \: \: { \bold{sum \: \: of \: \: roots = 36 + 8\: \: \:}}$

$\\ \: \: \: \: \: \: \: { \huge{.}} \: \: { \bold{sum \: \: of \: \: roots = 44\: \: \:}}$

$\\ \longrightarrow \: \: { \bold{product \: \: of \: \: roots \: \:of \: \: required \: \: \: ploynomial = { \alpha }^{2} . { \beta }^{2} }}$

$\\ \: \: \: \: \: \: \:{ \huge{.}} \: \: { \bold{product\: \: of \: \: roots = {( \alpha . \beta ) }^{2}}}$

$\\ \: \: \: \: \: \: \:{ \huge{.}} \: \: { \bold{product \: \: of \: \: roots = {( - 4) }^{2} \: \: \: \: \: \: [ \: using \: \: eq.(2) ]}}$

$\\ \: \: \: \: \: \: \:{ \huge{.}} \: \: { \bold{product \: \: of \: \: roots = 16\: \: \:}}$

• New quadratic equation –

$\\ \longrightarrow \: \: { \red { \bold{ {x}^{2} - (sum \: \: of \: \: roots)x + product \: \: of \: \: roots \: = 0\: \: \:}}}$

• So that required quadratic equation –

$\\ \longrightarrow \: \: \large { \green{ \boxed{ \bold{ {x}^{2} - 44x +16 = 0}}}}$