Question

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​

Answer 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

Answer 2

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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