Question

If the zeroes of the polynomial x2 + p x+ q are double in value to the zeroes of 2x2-5x -3, then Find the values of p and q.

Answer 1

⏺ Required Answer:

✏GiveN:

• Zeroes of x² + px + q are double of the zeroes of 2x² - 5x - 3

✏To FinD:

• Find the value of p and q....?

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⏺ How to solve...?

We have to know the relation between the zeroes of any polynomial and coefficients of its term. The relation is:

$\large{ \circ{ \boxed{ \rm{sum \: of \:zeroes = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}}} \\ \\ \large{ \circ{ \boxed{ \rm{product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}}}$

I know that many of us learn that, for any quadratic polynomial ax² + bx + c, Sum of zeroes = -b/a and product of series is c/a, the way is somewhat correct because the coefficients might change, and you can get confused.

✏ So, it's better to know the relation provided by me above!! Now we will solve this question, by using this relation.

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⏺ Solution:

Let $\alpha$ and $\beta$ be the zeroes of the polynomial 2x² - 5x - 3. Then, by relation,

$\large{ \rm{ \longrightarrow \: \alpha + \beta = - \frac{ (- 5)}{2}}} \\ \\ \large{ \rm{ \longrightarrow \: \alpha + \beta = \frac{5}{2}............(1)}} \\ \\ \large{ \underline{ \rm{ \red{and}}}} \\ \\ \large{ \rm{ \longrightarrow \: \alpha \beta = \frac{ - 3}{2}.............(2) }}$

✒ According to question,

Zeroes of polynomial x² + px + q will be $2\alpha$ and $2\beta$. Then, according to relation,

$\large{ \rm{ \longrightarrow \: 2 \alpha + 2 \beta = - \frac{p }{1} }} \\ \\ \large{ \rm{ \longrightarrow \: 2( \alpha + \beta ) = - p}} \\ \\ \large{ \rm{ \longrightarrow \: \alpha + \beta = \frac{ - p}{2} }}$

From, equation (1), we can say that,

$\large{ \rm{ \longrightarrow \: \frac{ - p}{2} = \frac{5}{2} }} \\ \\ \large{ \rm{ \longrightarrow \: - p = \frac{5 \times \cancel{2}}{ \cancel{2} }}} \\ \\ \large{ \rm{ \longrightarrow \: p = \boxed{ \red{ \rm{ - 5}}}}} \\ \\ \large{ \underline{ \red{ \rm{and}}}} \\ \\ \large{ \rm{ \longrightarrow \: 2\alpha \times 2 \beta = \frac{q}{1} }} \\ \\ \large{ \rm{ \longrightarrow \: \alpha \beta = \frac{q}{4} }}$

But from eq.(2), we can say that,

$\large{ \rm{ \longrightarrow \: \frac{q}{4} = \frac{ - 3}{2} }} \\ \\ \large{ \rm{ \longrightarrow \: q = \frac{ - 3 \times 4}{2} }} \\ \\ \large{ \rm{ \longrightarrow \: q = \boxed{ \rm{ \red{- 6}}}}}$

Hence the value of p = -5 and q = -6

$\large{ \therefore{ \underline{ \underline{ \green{ \rm{Hence \: solved \: \dag}}}}}}$

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