Question

If α and β are the roots of the equation x² + 5x + 5 =0 . Then Find the Equation whose roots are (α + 1) and (β + 1).​​

Answer 1

Topic

• Polynomials

A polynomial is

Solution

① The zeroes of the equation.

We are given an equation $x^{2}+5x+5=0$, and two zeroes are $\alpha$ and $\beta$. So, the roots of the equations satisfy the following.

❶ $\alpha ^{2}+5\alpha +5=0$

❷ $\beta ^{2}+5\beta +5=0$

Now we have the two remaining zeroes to find which will be $\alpha '$ and $\beta '$.

We can observe the following.

❶ $\alpha '=\alpha +1$

❷ $\beta '=\beta +1$

Then we also observe the following. Let's write in terms of the original zeroes.

❶ $\alpha =\alpha '-1$

❷ $\beta =\beta '-1$

② Finding the new equation.

Now the remaining zeros satisfy the following.

❶ $(\alpha '-1)^{2}+5(\alpha '-1)+5=0$

❷ $(\beta '-1)^{2}+5(\beta '-1)+5=0$

A new equation to be satisfied by new zeroes is the following.

$\implies (x-1)^{2}+5(x-1)+5=0$

$\implies (x^{2}-2x+1)+(5x-5)+5=0$

$\therefore x^{2}+3x+1=0$

This is the end of the required answer.

More Information

Consider a quadratic equation $ax^{2}+bx+c=0$. A quadratic equation that has reciprocal solutions is $cx^{2}+bx+a=0$.

In the similar method of the above, that it is true can be proved. This also works in a polynomial having a higher degree.

Solve More Questions

Question:-

Consider a biquadratic equation $ax^{4}+bx^{2}+c=0$. If the new equation has all reciprocal solutions of the given equation, which has the leading coefficient of $c$, on adding both equations, the equation always has _ as solutions. ($a+b+c$ is nonzero.)

① $\dfrac{1\pm \sqrt{3} i}{2}$

② $\dfrac{-1\pm \sqrt{3} i}{2}$

③ $\pm 1$

④ $0$

⑤ $\pm \sqrt{3}$

Answer: ②

Solution: New equation is $cx^{4}+bx^{2}+a=0$. When added we obtain $(a+b+c)(x^{2}+x+1)=0$. Hence ② is the correct option.

Answer 2

Given :-

If α and β are the roots of the equation x² + 5x + 5 = 0

To Find :-

The Equation whose roots are (α + 1) and (β + 1).​​

Solution :-

On comparing the given equation with ax² + bx + c.

We get

a = 1

b = 5

c = 5

We know that

Sum of zeroes = α + β = -b/a

Sum = -(5)/1

Sum = -5

Product of zeroes = αβ = c/a

Product = 5/1

Product = 5

Now,

New sum of zeroes = α + 1 + β + 1

Sum = α + 1 + β + 1

Sum = (α + β) + (1 + 1)

Sum = (-5) + 2

Sum = -3

New product of zeroes = (α + 1)(β + 1)

Product = (α + 1)(β + 1)

Product = (αβ) + (α + β) + 1

Product = 5 + (-5) + 1

Product = 1

Now,

Quadratic polynomial = x² - (α + β)x + αβ

Quadratic polynomial = x² - (-3)x + 1

Quadratic polynomial = x² + 3x + 1