Question

find the quadratic polynomial sum and product of whose zeroes are as given also find the zeroes if these polynomial by factorisation. (1) -2√3 ,-9​

Answer 1

EXPLANATION.

Quadratic polynomial whose sum and product are given,

As we know that,

Quadratic Polynomial ⇒ x² - (α + β)x + αβ.

Sum of zeroes of quadratic Equation,

⇒ α + β = -b/a.

⇒ α + β = -2√3.

Product of zeroes of quadratic Equation,

⇒ αβ = c/a.

⇒ αβ = -9.

Put the value in equation, we get.

⇒ x² - (α + β)x + αβ.

⇒ x² - (-2√3)x + (-9).

⇒ x² + 2√3x - 9.

⇒ x² + 2√3x - 9.

As we know that,

⇒ D = b² - 4ac.

⇒ D = (2√3)² - 4(1)(-9).

⇒ D = 12 + 36.

⇒ D = 48.

⇒ x = -b ± √D/2a.

⇒ x = -2√3 ± √48/2.

⇒ x = -2√3 - √48/2    and   x = -2√3 + √48/2.

                                                                                         

MORE INFORMATION.

Conjugate Roots,

If,

D < 0 ⇒ One Roots = α + iβ  other Roots = α - iβ.

D > 0 ⇒ One Roots = α + √β  other Roots = α - √β.

Answer 2

Lᴇᴛ :

☆ General form of the quadratic polynomial is

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\blue{x^2\:-\:(\alpha\:+\:\beta)\:x\:+\:\alpha\:\beta}}}}}}$

• α & β are the zeroes of the quadratic polynomial.

Gɪᴠᴇɴ :

• α + β = -2√3

• αβ = -9

Tᴏ Fɪɴᴅ :

• The zeroes of the quadratic polynomial.

Cᴀʟᴄᴜʟᴀᴛᴏʀ :

➣ Putting the value of (α + β) & αβ in the above equation, we get

$:\implies\:\:\bf{x^2\:-\:(-\:2\sqrt{3})\:x\:+\:(-\:9)}$

$:\implies\:\:\bf{x^2\:+\:\:2\sqrt{3}\:x\:-\:9}$

Fᴏʀᴍᴜʟᴀ ᴏғ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴘᴏʟʏɴᴏᴍɪᴀʟ :

$\pink\checkmark\:\:{\underline{\bf{\purple{x\:=\:\dfrac{-b\:\pm\:\sqrt{b^2\:-\:4ac}}{2a}\:}}}}$

[Nᴏᴛᴇ ➛ The symbol of ± indicates that there are two solutions of quadratic equation.]

$:\implies\:\:\bf{x\:=\:\dfrac{-b\:+\:\sqrt{b^2\:-\:4ac}}{2a}\:~~\&~~\:\dfrac{-b\:-\:\sqrt{b^2\:-\:4ac}}{2a}}$

Wʜᴇʀᴇ,

• a = 1

• b = 2√3

• c = -9

$:\implies\:\:\bf{x\:=\:\dfrac{-\:(2\sqrt{3})\:+\:\sqrt{(2\sqrt{3})^2\:-\:4.1.(-9)}}{2\times{1}}\:~\atop{\&~\:\dfrac{-\:(2\sqrt{3})\:-\:\sqrt{(2\sqrt{3})^2\:-\:4.1.(-9)}}{2\times{1}}}}$

$:\implies\:\:\bf{x\:=\:\dfrac{-\:(2\sqrt{3})\:+\:\sqrt{12\:+\:36}}{2}\:~\atop{\&~\:\dfrac{-\:(2\sqrt{3})\:-\:\sqrt{12\:+\:36}}{2}}}$

$:\implies\:\:\bf{x\:=\:\dfrac{-\:(2\sqrt{3})\:+\:\sqrt{48}}{2}\:~\atop{\&~\:\dfrac{-\:(2\sqrt{3})\:-\:\sqrt{48}}{2}}}$

$:\implies\:\:\bf{x\:=\:\dfrac{-\:(2\sqrt{3})\:+\:4\sqrt{3}}{2}\:~\atop{\&~\:\dfrac{-\:(2\sqrt{3})\:-\:4\sqrt{3}}{2}}}$

$:\implies\:\:\bf{x\:=\:\dfrac{2\sqrt{3}}{2}\:~~\&~~\:\dfrac{-\:6\sqrt{3}}{2}}$

$:\implies\:\:\bf\pink{x\:=\:\sqrt{3}\:~~\&~~\:-\:6\sqrt{3}}$

$\Large\bf\orange{Therefore,}$

The zeroes of the quadratic polynomial is √3 & -6√3.