Question

If α and β are zeroes of the polynomial q(x)=x2-7x+10, then find the value of (α+β)2-4αβ

Answer 1

Step-by-step explanation:

x²-7x+10

x²-5x-2x+10

x(x-5)-2(x-5)

(x-2)(x-5)

x=2 or x=5

alpha and beta are 2 and 5

(5+2)2-4(5)(2)

7(2)-4(10)

14-40

26

Answer 2

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✦ Required Answer:

♦️ GiveN:

• $\large{\alpha}$ and $\large{\beta}$ are the zeroes of the polynomial q(x) = $\large{\rm{{x}^{2} - 7x +10}}$.

♦️ To FinD:

• Value of $\large{\rm{{(\alpha + \beta)}^{2}-4 \alpha \beta}}$........?

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✦ Explanation of Concept:

The above question deals with the roots of the quadratic equation and we should know a simple relation between the zeroes to solve the above question.

Relation:-

If $\large{\alpha}$ and $\large{\beta}$ are the roots of the quadratic equation $\large{\rm{a{x}^{2}+ bx+c}}$

${ \dag{ \boxed{ \rm{ \purple{sum \: of \: the \: roots = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }}}}}} \\ \\ \large{ \rm{ \implies{ \alpha + \beta = \frac{ - b}{a} }}} \\ \\ { \dag{ \boxed{ \purple{ \rm{product \: of \: roots = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}}}} \\ \\ \large{ \rm{ \implies{ \alpha \beta = \frac{c}{a} }}}$

By using these formula or concepts, let's find the roots of the equation and the answer of the final sum.

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✦ Solution:

Given, equation $\large{\rm{{x}^{2} - 7x +10}}$.

And $\large{\alpha}$ and $\large{\beta}$ are its zeroes.

By using formula, Sum of roots

$\large{ \rm{ \dashrightarrow \: \alpha + \beta = \frac{ - ( - 7)}{1}}} \\ \\ \large{ \rm{ \dashrightarrow \: \alpha + \beta = 7 }}$

By using formula, Product of roots

$\large{ \rm{ \dashrightarrow \: \alpha \beta = \frac{10}{1}}} \\ \\ \large{ \rm{ \dashrightarrow \: \alpha \beta = 10}}$

We have to find:

$\large{ \star{ ( \alpha + \beta ) {}^{2} - 4 \alpha \beta }}$

Putting the above values,

$\large{ \rm{ \dashrightarrow \: {(7)}^{2} - 4(10)}} \\ \\ \large{ \rm{ \dashrightarrow \: 49 - 40}} \\ \\ \large{ \rm{ \dashrightarrow \: \boxed{ \red{ \rm{9 \: }}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \rm{ \blue{Hence \: solved \: \dag}}}}}}$

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