Question

If m and n are the zeroes of the polynomial x² + 7x + 7, then form a quadratic polynomial whose zeroes are 2m and 2n.​

Answer 1

Answer.

Given:-

Let $f(x)=x^{2}+7x+7$.

Roots of $f(x)$ are m and n.

Let the new zeros be m' and n', where

• $m'=2m$
• $n'=2n$

The solution:-

By definition of zeros, $f(m)=0$ and $f(n)=0$ is true.

If m and n are zeros of the polynomial, m' and n' will be the roots of $f\left(\dfrac{x}{2} \right )$, because we know that $f\left( \dfrac{m'}{2} \right) =0$ and $f\left( \dfrac{n'}{2} \right) =0$.

By above, $f\left(\dfrac{x}{2} \right)=k(\dfrac{1}{4} x^{2}+\dfrac{7}{2} x+7)$ has two zeros m' and n'.

We choose $k=4$ then the polynomial is $\boxed{x^{2}+14x+28}$.

Solve more questions.

Question example:-

Find the sum of multiplicative inverses of the zeros of $x^{2}+7x+7$.

What we need to know:-

A polynomial with its coefficients reversed has the inverses of the zeros. Here we are going to prove it.

The solution:-

Let α and β be the zeros of the equation. Then α' and β' will be new zeros.

Let's solve for $\alpha$ and $\beta$ in terms of $\alpha '$ and $\beta '$.

• $\alpha =\dfrac{1}{\alpha '}$
• $\beta =\dfrac{1}{\beta '}$

This means

• $f(\alpha )=0\implies f\left( \dfrac{1}{\alpha '} \right) =0$
• $f(\beta )=0\implies f\left( \dfrac{1}{\beta '} \right) =0$

So, $f\left( \dfrac{1}{x} \right)$ has $\alpha '$ and $\beta '$ as zeros.

$\implies f\left(\dfrac{1}{x} \right) =7\left( \dfrac{1}{x} \right) ^{2}+7\left( \dfrac{1}{x} \right)+1$

$\implies f\left(\dfrac{1}{x} \right) = \dfrac{7}{x^{2}} ^{2}+\dfrac{7}{x} +1$

Multiplying $x^{2}$ on both sides,

$x^{2}\cdot f\left(\dfrac{1}{x} \right) =7+7x+x^{2}$

We get $x^{2}+7x+7$ as a polynomial whose zeros are the inverses.

The sum of the inverses of the zero is hence $-\dfrac{7}{7} =\boxed{-1}$.

Answer 2

Given :-

If m and n are the zeroes of the polynomial x² + 7x + 7

To Find :-

Quadratic polynomial

Solution :-

Here

α + β = m + n

m + n = -b/a

Where

b = 7

a = 1

m + n = -(7)/1

m + n = -7

Product of zeroes = αβ = mn

mn = c/a

c = 7

a = 1

mn = 7/1

mn = 7

Now

New sum of the zeroes = 2m + 2n = 2(m + n) = 2(-7) = -14

New product of zeroes = 2m × 2n = 4(m × n) = 4(7) = 28

Now

We know that

Polynomial = x² - (α + β)x + αβ

Polynomial = x² - (-14)x + 28

Polynomial = x² + 14x + 28

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