Question

If a and b are the zeroes of the polynomial x^2 -5x +4, find a quadratic polynomial having zeroes a+1/b and b+1/a

Answer 1

Solution

Consider a and b to be the zeros of the polynomial,p(x) = x² - 5x + 4

Here,

a + b = 5............[1]

and, ab = 4................[2]

Now,

a² + b² = (a + b)² - 2ab

→a² + b² = (5)² -2(4)

→a² + b² = 25 - 8

→a² + b² = 17............[3]

• Let S and P denote the sum and product of the zeros of the required polynomial

Now,

S = (a+1)/b + (b+1)/a

→S = [a(a+1)+b(b+1)]/ab

→S = [(a²+b²)+(a+b)]/ab

→S = (17+5)/4

→S = 22/4

→S = 11/2

Also,

P = [(a+1)/a][(b+1)/b]

→ P = [(a+1)(b+1)]/ab

→P = [(a+b)+ab+1]/ab

→P = (5+4+1)/4

→P = 10/4

→P = 5/2

★Required Polynomial is of the form:

x² - Sx + P

= x² -(11/2)x + 5/2

•Dividing the equation throughout by 2,we get:

= 2x² - 11x + 5

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Quadratic\:eqn=4x^{2}-25x+25}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a quadratic whose zeroes are a and b are the zeroes of the polynomial x^2 -5x +4.

• We have to find a quadratic polynomial having zeroes a+1/b and b+1/a.

$\underline \bold{Given : } \\ \implies a \: \: and \: \: b \in \: ( {x}^{2} - 5x + 4) \\ \\ \underline \bold{to \: find : } \\ \implies Quadratic\:eqn\:form\:by\:given\:zeroes= ?$

• According to given question :

$\bold{By \: Quadratic \:formula : } \\ \implies {x}^{2} - 5x + 4 = 0 \\ \\ \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \implies x = \frac{ - ( - 5) \pm \sqrt{ { (- 5)}^{2} - 4 \times 1 \times 4 } }{2 \times 1} \\ \\ \implies x = \frac{5 \pm \sqrt{25 - 16} }{2} \\ \\ \implies x = \frac{5 \pm3}{2} \\ \\ \bold{\implies x = 4 \: and \:1 }\\ \\ \therefore \bold{a = 4} \\ \\ \therefore \bold{b = 1} \\ \\ \bold{For \: sum \: of \: zeroes : } \\ \implies a + \frac{1}{b} + b + \frac{1}{a} \\ \\ \implies \frac{ {a}^{2}b + a + {ab}^{2} + b }{ab} \\ \\ \implies \frac{ {4}^{2} \times 1 + 4 + 4 \times {1}^{2} + 1 }{4 \times 1} \\ \\ \implies \frac{16 + 4 + 4 + 1}{4} \\ \\ \bold{\implies \frac{25}{4} } \\ \\ \bold{ For\:product \: of \: zeroes : } \\ \implies (a + \frac{1}{b} ) \times (b + \frac{1}{a} ) \\ \\ \implies \frac{ab + 1}{b} \times \frac{ab + 1}{a} \\ \\ \implies \frac{4 \times 1 + 1}{1} \times \frac{4 \times 1 + 1}{4} \\ \\ \implies \frac{5 \times 5}{4} \\ \\ \bold{\implies \frac{25}{4} } \\ \\ \bold{For \: Quadratic\:equation : } \\ \implies {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0 \\ \\ \implies {x}^{2} - \frac{25x}{4} + \frac{25}{4} = 0 \\ \\\implies \frac{4x^{2}-25x+25}{4}=0\\\\ \bold{\implies {4x}^{2} - 25x + 25 = 0}$