Question

4. Find the quadratic polynomial if its zeroes are 0, √5.

5. Find the value of “x” in the polynomial 2a² + 2xa + 5a + 10 if (a + x) is one of its factors.

6. How many zeros does the polynomial (x – 3)² – 4 can have? Also, find its zeroes.​

Answer 1

Step by step explanation :-

4) Given zeroes of the Quadratic polynomial 0, √5.

We have to find the Quadratic polynomial

If ${\alpha,\beta}$ are the roots of Quadratic polynomial then the Quadratic polynomial is

$x {}^{2} - ( \alpha + \beta )x + \alpha \beta$

$\alpha = 0$

$\beta = \sqrt{5}$

$\alpha + \beta = 0 + \sqrt{5}$

$\alpha + \beta = \sqrt{5} \: \: eq1$

$\alpha \beta = 0( \sqrt{5} )$

$\alpha \beta = 0 \: eq2$

Substituting in formula

$x {}^{2} - ( \alpha + \beta )x + \alpha \beta$

$x {}^{2} - \sqrt{5} x + 0$

$x {}^{2} - \sqrt{5} x$

This is the required polynomial

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5) We have to find the value of 2a² + 2xa + 5a + 10

One of factor is a+x

a + x = 0

a = - x Substituting value in above equation

2a² + 2xa + 5a + 10

2(-x)² + 2(x)(-x) +5(-x) + 10 = 0

2x² -2x² -5x + 10 = 0

-5x + 10 = 0

-5x = -10

5x = 10

x = 10/5

x = 2

So, the value of x is 2

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6) We have to find how many zeroes are there for above polynomial and what are the zeroes

(x – 3)² – 4

We shall expand this

x² + 9 - 6x - 4

x² - 6x + 5

Hence Its degree is 2 So, It is a Quadratic polynomial

Since, Quadratic polynomial has 2 zeroes

So, it has 2 zeroes

Now finding the zeroes

x² - 6x + 5 = 0

Splitting the middle term

x² -x - 5x + 5 = 0

x(x -1) -5 (x -1) = 0

(x -1) (x -5) = 0

x-1 = 0

x = 1

x-5 = 0

x = 5

So, the zeroes are 1, 5