Question

1. Find the value of “p” from the polynomial x² + 3x + p, if one of the zeroes of the polynomial is 2.

2. Does the polynomial a⁴ + 4a² + 5 have real zeroes?

3.Compute the zeroes of the polynomial 4x² – 4x – 8. Also, establish a relationship between the zeroes and coefficients.​

Answer 1

1. let x = 2 , p(x) = x² + 3x + p = 0

• (4)² + 3(4) + p = 0
• 16 + 12 + p = 0
• p = -28
• Value of p = -28

2. No

Answer 2

Answer:

Step-by-step explanation:

Solution :-

1) Given , value of 'x' = 2

So by substituting the value of 'x' and equating it to zero we will get the value of 'p'

Lets do :-

$(2)^{2}+3(2)+p = 0$

$4+6+p=0$

10+p = 0

p = -10

Since, Value of 'p' = -10

2) $a^{4}+4a^{2}+5$

$=( a^{2})^{2}+4a^{2} + 5$

$Let, a^{2}=x$

$\implies x^{2}+4a+5$

Comparing it with general form of quadratic equation "ax^2+bx+c"

We get, a = 1, b = 4 , c = 5

If we get$\triangle=b^2-4ac<0\implies$The quadratic equation have no real roots

$b^2-4ac=(4)^2-4(1)(5)$

$= 16 - 20$

= - 4

$\therefore \triangle = -4<0\implies$the quadratic equation has no real roots.

3) $4x^2-4x-8$

To Find the zeroes of the polynomial we need to equate the polynomial to zero :-

$4x^2-4x-8=0$

$4(x^2-x-2)=0$

$x^2-x-2=0$

$x^2-2x+x-2=0$

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

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

x - 2 = 0 and x + 1 = 0

x = 2 and x = -1

$\implies$Values of x = 2,-1.

Consider a general quadratic equation "$ax^2+bx+c=0$" and roots of the quadratic equation as "$\alpha ,\beta$"

$\implies\alpha +\beta = \dfrac{-b}{a}$

$\alpha \times\beta = \dfrac{c}{a}$