Question

if a and b are the zeroes of the polynomial 2x2 +6x -3, find the value of a2 +b2​

Answer 1

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Answer

• α² + β² = 12

$\orange{\bigstar}$  Given  $\green{\bigstar}$

α and β are the zeroes of the polynomial 2x² + 6x - 3

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

Value of a² + b²

$\orange{\bigstar}$  Key points  $\green{\bigstar}$

For a quadratic polynomial ax² + bx + c ,

Sum of zeroes ,

$\bf \alpha +\beta=\dfrac{-b}{a}$

Product of zeroes ,

$\bf \alpha \beta=\dfrac{c}{a}$

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→ ( x + y )² - 2xy = x² + y²

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

Compare given equation 2x² + 6x - 3 with ax² + bx + c ,

⇒ a = 2  ,  b = 6  ,  c = - 3

Now ,

Sum of zeroes ,

$\rm \alpha +\beta =\dfrac{-b}{a}\\\\\to \rm \alpha + \beta=\dfrac{-6}{2}\\\\\to \rm \alpha + \beta =-3\ ...(1)$

Product of zeroes ,

$\rm \alpha \beta=\dfrac{c}{a}\\\\\to \rm \alpha \beta=-\dfrac{3}{2}...(2)$

Now , Let's find our required value ,

$\to \rm \alpha^2+\beta^2\\\\\to \rm (\alpha+\beta)^2-2\alpha \beta\\\\\to \rm (-3)^2-2\left( -\dfrac{3}{2}\right)\\\\\to \rm 9+3\\\\\to \bf 12\ \; \pink{\bigstar}$

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Answer 2

The value of α² + β² = 12

Step-by-step explanation:

★ Given that :

• α and β are the zeroes of the polynomial p(x) = 2x² + 6x - 3

★ To find :

• Value of α² + β².

★ Let :

◼ The general form of a Quadratic Polynomial is ax² + bx + c = 0.

◼ Consider the given polynomial p(x).

• a = 2
• b = 6
• c = - 3

★ Now :

$\bf\red{\rightarrow Sum\:of\:the\:zeroes : \alpha +\beta = \cfrac{-b}{a} }$

$\sf \implies \alpha + \beta = \cfrac{-6}{2}$

$\sf \implies \alpha + \beta = -3$

$\bf\red{\rightarrow Product\:of\:the\:zeroes : \alpha \beta = \cfrac{c}{a} }$

$\sf \implies \alpha\beta = \cfrac{-3}{2}$

We know that identity,

(a + b)² = a² + b² + 2ab.

Now, consider it as

(α + β)² = α² + β² + 2αβ

• Substitute the zeroes.

$\sf \implies (-3)^{2} = \alpha^{2} + \beta^{2} + \cancel{2}(\cfrac{-3}{\cancel{2}})$

$\sf \implies 9 = \alpha^{2} + \beta^{2} - 3$

$\sf \implies 9 + 3= \alpha^{2} + \beta^{2}$

$\sf \implies \alpha^{2} + \beta^{2}=12$

$\underline{\boxed{\rm{\purple{\therefore Value\:of\: \alpha^{2} + \beta^{2} = 12.}}}}\:\orange{\bigstar}$