Math, asked by ravichandranha66, 9 months ago

4. If a and ß are the zeroes of the quadratic polynomial f(x) = x² – 3x - 2, then
find a quadratic polynomial whose zeroes are air and 26+a
e Dan aanh of the following find a quadratic polynomial whose sum and product are 1/2a+b and 1/2b+a​

Answers

Answered by nprabakaran32
1

Answer:

x2-3x-2=0

2x2

x2-2x-x-2=0

(x2-2x) (-x-2)=0

x(x-2)-1(x-2)=0

(x-2)(x-1)=0

x-2=0. x-1=0

so, x=2. x=1

so the value of x is 2 and 1

Answered by sonuvuce
0

The required quadratic polynomial is 8x²-18x-9

Step-by-step explanation:

If \alpha and \beta are the zeroes of quadratic polynomial

f(x)=x^2-3x-2 then

\alpha+\beta=-\frac{\text{Coefficient of x}}{\text{Coefficient of x^2}}

\implies \alpha+\beta=-(\frac{-3}{1})=3

Also,

\alpha\beta=\frac{\text{Constant Term}}{\text{Coefficient of x^2}}

\implies \alpha\beta=\frac{-2}{1}=-2

We have to find the quadratic polynomial whose zeroes are \alpha+\frac{1}{2\beta} and \beta+\frac{1}{2\alpha}

Sum of zeroes

=\alpha+\frac{1}{2\beta}+\beta+\frac{1}{2\alpha}

=(\alpha+\beta)+\frac{1}{2}(\frac{1}{\alpha}+\frac{1}{\beta}

=3+\frac{\alpha+\beta}{2\alpha\beta}

=3+\frac{3}{2\times(-2)}

=3-\frac{3}{4}

=\frac{9}{4}

Product of zeroes

=(\alpha+\frac{1}{2\beta})\times(\beta+\frac{1}{2\alpha})

=\alpha\beta+\frac{1}{2}+\frac{1}{2}+\frac{1}{4\alpha\beta}

=-2+1+\frac{1}{4\times(-2)}

=-1-\frac{1}{8}

=-\frac{9}{8}

We know that quadratic polynomial whose zeroes are \alpha and \beta is given by

x^2-(\alpha+\beta)x+\alpha\beta

Therefore, the quadratic polynomial is

x^2-(\frac{9}{4})x+(-\frac{9}{8})

or, x^2-\frac{18}{8}x-\frac{9}{8}

or, 8x^2-18x-9

Hope this answer is helpful.

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