Question

12. If a and B are the zeroes of the quadratic polynomial f(x) = 2x? – 5x + 7, then find a quadratic
polynomial whose zeroes are 2a +3ß and 2B +3a.
​

Answer 1

Answer:

If A and B are the zeros of the polynomial,p(x) = 2x² - 5x + 7. Then find a quadratic polynomial whose zeros are "2A + 3B" and "2B + 3A".

Given that A and B are the zeros of the polynomial 2x² - 5x + 7

Note

Sum of Zeros : - x coefficient/x² coefficient

Product of Zeros : constant term/x² coefficient

Here,

Sum of Zeros :

A + B = -(-5)/2

→ A + B = 5/2

Product of Zeros :

AB = 7/2

Let S and P be the sum and product of zeros of required polynomial

Sum of Zeros

S = (2A + 3B) + (3A + 2B)

→ S = 5(A + B)

→ S = 5(5/2)

→ S = 25/2

Product of Zeros

P = (2A + 3B)(3A + 2B)

→ P = 6A² + 4AB + 9AB + 6B²

→ P = 6A² + 13AB + 6B²

→ P = (6A² + 12AB + 6B²) + AB

→ P = 6(A + B)² + AB

→ P = 6(5/2)² + 7/2

→ P = 150/4 + 7/2

→ P = 75/2 + 7/2

→ P = 82/2

Required Polynomial

x² - Sx + P

= x² - (25/2)x +82/2

= 2x² - 25x + 82

Thus,the required quadratic polynomial is 2x² - 25x + 82

It might help you ❤❤

Answer 2

$\Huge\bf\underbrace{\underline{\blue \dag \green{Question}}}$

If A and B are the zeros of the quadratic polynomial f(x) = 2x² - 5x + 7 . then find a quadratic whose zeros are (2A+3B) and ( 2B + 3A) .

$\huge\mathbb\fcolorbox{Green}{violet}{♡ᎪղՏωᎬя᭄}$

$\to \boxed{ \sf {x}^{2} - ( \frac{25}{2} )x + 41 = 0}$

$\begin{gathered}\Large\bf{\color{lime}To FiNd,} \\ \end{gathered}$

→ Another Quadratic polynomial.

$\begin{gathered}\bf\blue{Explanation :-,} \\ \end{gathered}$

We have ,

$\begin{gathered} \leadsto \sf f(x) = 2 {x}^{2} - 5x + 7 \\ \\ \sf \to sum \: of \: roots(A + B) = \frac{ - ( - 5)}{2} \\ \\ \to \sf A + B = \frac{5}{2} \: \: \: ...(1) \\ \\ \to \sf product \: of \: roots (A B) = \frac{7}{2} ..(2) \\ \end{gathered}$

Now,

We know that ,

we have required quadratic polynomial -

• → ♦x² -(s)x +p= 0
• Its roots are (2A + 3B ) and (3A + 2B)
• → Sum of roots = 2A + 3B + 3A + 2B

$\begin{gathered} \to \sf sum \: of \: roots(s) = 5(A + B) \: \: \\ \sf from \: (1) \\ \\ \to \sf \:(s) = 5 \times \frac{5}{2} \\ \\ \to \sf \boxed{ \sf (s) = \frac{25}{2} } \\ \\ \leadsto \sf product \: of \: roots(p) = (2A + 3B)(3A + 2B) \\ \\ \leadsto \sf \:(p) = 6{A }^{2} + 4A B + 9A B + 6 { B}^{2} \\ \\ \leadsto \sf (p) = 6{A }^{2} + 6 { B}^{2} + 13A B \\ \\ \leadsto \sf (p )= 6({A }^{2} \\ \end{gathered}$

                           $+ { B}^{2} + 2A B) + A B \: \\ \\ \leadsto \sf \: (p )= 6{(A + B)}^{2} + A B \\ \\ \leadsto \sf (p )= 6 \times \frac{25}{4} + \frac{7}{2} \\ \\ \leadsto \sf (p) = \frac{75}{2} + \frac{7}{2} \\ \\ \leadsto \sf(p) = \frac{82}{2} = 41 \\ \end{gathered}$

Now , required quadratic polynomial is -

$\to \boxed{ \sf {x}^{2} - ( \frac{25}{2} )x + 41 = 0}$