Question

Q1. Find zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients. (The zeroes are a, 1/a)

$g(x) = a(x^2+1) - x(a^2+1)$

_______________________________

Q2. If one zero of the following polynomial is negative of the other, find the value of k. (The answer is k = 0).

$4x^2 - 8kx - 9$

_______________________________

Q3. If the squared difference of the zeroes of the following quadratic polynomial is 144, find the value of p. (The answer is p = ± 18)

$f(x) = x^2 + px + 45$

Thanks!

Answer 1

Question :- 1 :)

Answer :-

→ 1/a or a .

Step-by-step explanation :-

We have,

→ A quadratic polynomial :

→ a( x² + 1 ) - x( a² + 1 ) = 0 .

==> ax² + a - x( a² + 1 ) = 0 .

==> ax² - ( a² + 1 )x + a = 0 .

Here, A = a , B = -( a² + 1 ) and C = a .

==> ax² - a²x - x + a = 0 .

==> ax( x - a ) - 1( x - a ) = 0 .

==> ( ax - 1 ) ( x - a ) = 0 .

==> ax - 1 = 0 or x - a = 0 .

•°• x = 1/a or a .

VERIFICATION :)

Therefore, Sum of zeros = -( coefficient of x )/( coefficient of x² ) = - B/A

==> 1/a + a = -( -( a² + 1 ) )/ a .

•°• ( 1 + a² )/a = ( 1 + a² )/ a .

And, Product of zeros = Constant term/ coefficient of x² = C/A .

==> 1/a × a = a/a .

•°• 1 = 1 .

Question :- 2 :)

Answer :-

→ k = 0 .

Step-by-step explanation :-

It is given that,

→ One zeros of the given polynomial is negative of the other .

Let one zero of the given polynomial be x .

Then, the other zero is -x .

•°• Sum of zeros = x + ( - x ) = 0 .

But, Sum of zeros = -( coefficient of x )/( coefficient of x² ) = - ( -8k )/4 .

==> 2k = 0 .

==> k = 0/2 .

•°• k = 0 .

Question :- 3 :)

Answer :-

→ p = ± 18 .

Step-by-step explanation:-

It is given that ,

→ The squared difference of the zeroes of the given quadratic polynomial is 144 .

°•° ( α - β )² = 144 .

Let α and β are the two zeros of the given polynomial .

The given quadratic polynomial is f(x) = x² + px + 45 .

we have, α+ β= -p ( Sum of zeros )

and , αβ = 45 ( Products of zeros ) .

Now,

°•° (α - β)² = 144 .

==> (α + β)² - 4αβ = 144

we have (α+β) = -p and αβ = 44 then put it's value in given equation

So,

==> p² - 4 × 45 = 144

==> p ² - 180 = 144

==> p² = 144+ 180

==> p² = 324

==> p = √324

$\therefore$ p = ±18 .

Hence, it is solved .

Answer 2

Q1)

$g(x) = a(x^2+1) - x(a^2+1)$

$g(x) =ax^{2}+a-a^{2}x-x \\ \\ \\g(x)=ax^{2}-a^{2}-x+a\\ \\ \\g(x)=ax(x-a)-1(x-a)\\ \\ \\g(x)=(x-a)(ax-1)$

also,

$g(x) =ax^{2}+a-a^{2}x-x\\ \\ \\g(x)=ax^{2}-x(a^{2}+1) +a$

Zeroes are as follows:-

$(x-a)=0\\ \\ \\x=a \\ \\ \rule{200}{2} \\ (ax-1)=0\\ \\ \\x= \frac{1}{a}$

$\rule{100}{2}$

Sum of zeroes = $a+\frac{1}{a} = \frac{a^{2}+1}{a}$

$\frac{-b}{a}=\frac{a^{2}+1}{a}$

hence,

sum of Zeroes = $\frac{-b}{a}$

product of zeroes=$a \times \frac{1}{a}= 1$

$\frac{c}{a}=\frac{a}{a}=1$

hence,

product of zeroes=$\frac{c}{a}$

$\rule{200}{2}$

Q2)

$4x^2 - 8kx - 9$

Let one Zero be a and other zero be (-a)

hence,

sum of zeroes= a+(-a)=0

so,

$\frac{8k}{4}=0 \\ \\ \\ \frac{2k}{1}=0 \\ \\ \\ k=0$

$\rule{200}{2}$

Q3)

$f(x) = x^2 + px + 45$

Let the zeroes be $\alpha\:and\: \beta$

As per question:-

$(\alpha-\beta) ^{2} =144\\ \\ \\ \alpha ^{2} +\beta ^{2}-2\alpha \beta =144\\ \\ \\(\alpha +\beta) ^{2}-4\alpha \beta =144\\ \\ \\ (-p)^{2}-4(45)=144\\ \\ \\ p^{2}=144+180=324\\ \\ \\ p= \sqrt{324} =\pm18$

$\red{\rule{200}{2}}$