Question

If f(x) be a quadratic polynomial such that f(x) = 0 has a root 3 and f(2) + f(-1) = 0 then other root lies in

(1) (-1,0)
(2) (0.1)
(3) (-2,1)
(4) (1.2)

Answer 1

ANSWER:

• Option 1) (- 1 , 0) is correct.

GIVEN:

• f(x) is a quadratic polynomial such that f(x) = 0.

• One of the root is 3 .

• f(2) + f(- 1) = 0.

TO FIND:

• The interval of other root.

EXPLANATION:

f(x) = x² - (sum of roots)x + product of roots

One root = 3

Let the other root be k.

Sum of roots = 3 + k

Product of roots = 3k

f(x) = x² - (3 + k)x + 3k = 0

f(2) + f(- 1) = 0.

f(2) = 2² - (3 + k)2 + 3k

f(2) = 4 - (6 + 2k) + 3k

f(2) = 4 - 6 - 2k + 3k

f(2) = - 2 + k

f(- 1) = (- 1)² - (3 + k)(- 1) + 3k

f(- 1) = 1 + (3 + k) + 3k

f(- 1) = 1 + 3 + k + 3k

f(- 1) = 4 + 4k

f(2) + f(- 1) = - 2 + k + 4 + 4k = 0

2 + 5k = 0

5k = - 2

k = - 0.4

- 1 < - 0.4 < 0

- 0.4 ∈ (- 1 , 0)

HENCE - 0.4. LIES IN THE INTERVAL (- 1 , 0).

Answer 2

GIVEN:-

• f(x) is a quadratic polynomial such that f(x) = 0 has a root 3 and f(2) + f(-1) = 0 .

TO FIND :-

• interval of other root .

CONCEPT:-

f(x) = x^2 - ( $\alpha$ + $\beta$ )x + $\alpha \beta$

Where ,

$\alpha \: and \: \beta \: are \: two \: roots$

SOLUTION:-

Given that one root is 3 .

Assume another root is K.

Then,

($\alpha$ + $\beta$) = 3 + k

$\implies \: \alpha \beta$ = 3k

$f(x) = x {}^{2} - (3 + k)x + 3k = 0$

Given that ,

f(2) + f( - 1) = 0

$\implies \: f(2) = 2 {}^{2} - (3 + k)2 + 3k \\ \\ \implies \: 4 - 6 - 2k + 3k \\ \\ \implies \: f(2) = - 2 + k$

Also,

$\\ \implies \: f( - 1) = ( - 1) {}^{2} - (3 + k)( - 1) + 3k \\ \\ \implies \: 1 + 3 + k + 3k \\ \\ \implies \: f( - 1) = 4 + 4k \\ \\ \implies \: f(2) + f( - 1) = - 2 + k + 4 + 4k = 0 \\ \\ \implies \: 2 + 5k = 0 \\ \\ \implies \: k = \frac{ - 2}{5} . \\ \\ \rightarrow \: - 1 < \frac{ - 2}{5} < 0 \\ \\ \implies \: \frac{ - 2}{5} \in \: ( - 1 ,0).$

Hence, the answer is (-1,0)

HOPE IT HELPS :)