Question

quadratic polynomial 4x²+12x+9 has zeroes as alpha and bita is now form a quadratic polynomial whose zeroes are alpha-1 and bita-1​

Answer 1

Answer:

Hear giving equation 4x²+12x+9

then zeroes of polynomial write

(x+1)(x+1)=x²+1

after all zeroes polynomial division by equation

then you can easily answer

Answer 2

Given

• $\bf \: 4 {x}^{2} + 12x + 9 \: has \: zeroes \: as \: \alpha \: and \: \beta$

To Find

• $\bf \: quadratic \: polynomial \: whose \: zeroes \: are \: \alpha - 1 \: and \beta - 1$

Concept used

• $\bf \: \alpha + \beta = \dfrac{ - b}{a}$

• $\bf \: \alpha \beta = \dfrac{c}{a}$
• $\bf \: required \: equation \\ \bf k\{{x}^{2} - (s)x + (p) \}$

Solution

For quadratic equation

$\bf \: 4 {x}^{2} + 12x + 9$

• a=4
• b=12
• c=9

$\bf \: \alpha + \beta = \dfrac{ - b}{a} \\ \\ \implies \boxed{ \bf \alpha + \beta = \frac{ - 12}{4}} \\ \\ \bf \: and \: \alpha \beta = \dfrac{c}{a} \\ \\ \implies \boxed{ \bf \alpha \beta = \dfrac{9}{4} } \: \: \: \: \: \: \: \: \: \: \: \:$

For required quadratic equations

$\bf \: sum(s) = \alpha - 1 + \beta - 1 \\ \\ \implies \bf \: sum(s) = \dfrac{ - 12}{4} - 2 \\ \\ \implies \bf sum(s) = \dfrac{ - 12 - 8}{4} \\ \\ \implies \boxed{\bf sum(s) = \dfrac{ - 20}{4} } \: \: \: \: \: \\ \\ \\ \\ \bf \: product(p) = ( \alpha - 1)( \beta - 1) \\ \\ \bf \implies product(p) = \alpha \beta - ( \alpha + \beta ) + 1 \\ \\ \implies \bf product(p) = \dfrac{9}{4} + \dfrac{12}{4} + 1 \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \: product(p) = \dfrac{9 + 12 + 4}{4} \: \: \: \: \: \: \: \: \\ \\ \bf \implies \boxed{ \bf \: product(p) = \dfrac{25}{4} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Required equation

$\bf \: k( {x}^{2} - (s)x + (p) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \: k( {x}^{2} + \dfrac{20}{4} + \frac{25}{4} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \: k( \dfrac{4 {x}^{2} + 20 + 25 }{4} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \underline{ \boxed{ \huge \bf 4 {x}^{2} + 20 + 25 }} \bf \: is \: the \: required \: equation \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: where \: k = 4$