Math, asked by mkj54, 1 year ago

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Answered by Anonymous

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$\large\mathcal\red{solution}$

Here I'm taking(alpha)=a

and (beta)=b

The given equation is ...

x²-2x-8=0

Now ...a and b are the zeros so....

a+b=2

And ....ab=-8

Now.....

(1)

A quadratic equation with ,...zeroes 2a and 2b is ,....

x²-(2a+2b)X+(2a×3b)=0

Now...(2a+2b)=2(a+b)=2×2=4

And ..(2a)(2b)=4(ab)=4(-8)=-32

.... therefore the equation

X²-4x-32=0

(2)

A quadratic equation with ,...zeroes 3a and 3b is ,....

x²-(3a+3b)X+(3a×3b)=0

Now...(3a+3b)=3(a+b)=3×2=6

And ..(3a)(3b)=9(ab)=9(-8)=-72

.... therefore the equation

X²-6x-72=0

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$

Answered by Anonymous

$\huge\bigstar\mathfrak\red{\underline{\underline{SOLUTION}}}$

${x}^{2} - 2x - 8 = 0 \\ = > \alpha + \beta = - ( - 2) = 2 \\ \alpha \beta = - 8 \\ \\ = > 2 \alpha + 2 \beta = 2( \alpha + \beta ) = 2(2) = 4 \\ \\ = > 2 \alpha 2 \beta = 4 \alpha \beta = 4( - 8) = - 32$

Required quadratic polynomial,

${x}^{2} - (2 \alpha + 2 \beta )x + (2 \alpha )(2 \beta ) = 0 \\ = > {x}^{2} - 4x - 32 = 0$

hoep it helps ☺️

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Required quadratic polynomial,

hoep it helps ☺️

Plz answer if u know ...