Question

If alpha and beta are thr roots of the equation 2x^2 +3x +2 =0 ,find the equation whose roots are alpha +1 and beta +1

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If α and β are the roots of the equation 2x² + 3x + 2 = 0

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The equation whose roots are α + 1 and β + 1.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We have quadratic equation as compared with ax² + bx + c

• a = 2
• b = 3
• c = 2

Now;

$\underline{\underline{\bf{Sum\;of\:the\:zeroes\::}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:(x)^{2} }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{\green{\alpha +\beta =\dfrac{-3}{2} }}$

$\underline{\underline{\bf{Product\:of\:the\:zeroes\::}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:(x)} }\\\\\\\mapsto\sf{\alpha \times \beta =\cancel{\dfrac{2}{2} }}\\\\\\\mapsto\sf{\green{\alpha \times \beta =1}}$

So;

We have the roots of sum of the roots :

$\mapsto\sf{\alpha +1+\beta +1}\\\\\\\mapsto\sf{\alpha +\beta +2}\\\\\\\mapsto\sf{\dfrac{-3}{2} +2\:\:\:\:\:\:\bigg[\therefore \alpha +\beta =\dfrac{-3}{2} \bigg]}\\\\\\\mapsto\sf{\dfrac{-3+4}{2} }\\\\\\\mapsto\sf{\green{\dfrac{1}{2} }}$

We have the roots of product of the roots :

$\mapsto\sf{(\alpha +1)(\beta+1)}\\\\\\\mapsto\sf{\alpha \beta +\alpha +\beta +1}\\\\\\\mapsto\sf{1+\dfrac{-3}{2} +1\:\:\:\:\:[\therefore \alpha \beta =1]}\\\\\\\mapsto\sf{2+\dfrac{-3}{2}} \\\\\\\mapsto\sf{\dfrac{4-3}{2} }\\\\\\\mapsto\sf{\green{\dfrac{1}{2} }}$

Thus;

The quadratic equation are required :

$\longrightarrow\sf{x^{2} -(sum\:of\:roots)x+(product\:of\:roots)}\\\\\longrightarrow\sf{x^{2} -\bigg(\dfrac{1}{2} \bigg)x+\bigg(\dfrac{1}{2} \bigg)}\\\\\longrightarrow\sf{x^{2} -\dfrac{1}{2}x +\dfrac{1}{2} }\\\\\longrightarrow\sf{\red{2x^{2} -x+1}}$

Answer 2

The quadratic equation is $\bf 2 {x}^{2} - x + 1 = 0$.

Step-by-step explanation:

Given:

• If $\alpha$ and $\beta$ are the roots of the equation $2 {x}^{2} + 3x + 2 = 0$.

To find:

• Find the equation whose roots are $\alpha+1$ and $\beta+1$.

Solution:

Concept to be used:

• The standard quadratic equation is given as $\bf a {x}^{2} + bx + c = 0$, here $a \neq0$and if $\alpha$ and $\beta$ are the roots of equation then

1. $\alpha + \beta = \frac{ - b}{a}$ and
2. $\alpha \beta = \frac{c}{a}$

• The quadratic equation in terms of zeros can be written as $\bf {x}^{2} - ( \alpha +\beta)x + ( \alpha\beta) = 0$.

Step 1:

Write the coefficients of equation.

Compare the given quadratic equation with standard one and write the coefficients of x², x, and constant term.

$\bf a = 2$

$\bf b = 3$

and

$\bf c = 2$

Step 2:

Write the relationship between zeros and coefficients.

As $\alpha$ and $\beta$ are the roots of the equation.

So,

$\bf \alpha + \beta = \frac{ - 3}{2}...eq1$

and

$\bf \alpha \beta = \frac{2}{2} = 1...eq2$

Step 3:

Calculate the values of relationship of roots for new equation.

As roots are $\alpha+1$ and $\beta+1$.

So,

Add +2 in both sides of eq1.

$(\alpha + 1) + ( \beta + 1) = \frac{ - 3}{2} + 2$

or

$(\alpha + 1) + ( \beta + 1) = \frac{ - 3 + 4}{2}$

or

$\bf (\alpha + 1) + ( \beta + 1) = \frac{ 1}{2} ...eq3$

and

$( \alpha + 1)( \beta + 1) = \alpha \beta + \alpha + \beta + 1$

put values from eq1 and eq2.

$( \alpha + 1)( \beta + 1) = 1 - \frac{3}{2} + 1$

or

$( \alpha + 1)( \beta + 1) = \frac{2 - 3 + 2}{2}$

or

$\bf ( \alpha + 1)( \beta + 1) = \frac{1}{2} ...eq4$

Step 4:

Write the quadratic equation.

${x}^{2} - ( \alpha + 1 + \beta + 1)x + ( \alpha + 1)( \beta + 1) = 0$

put the values from eq3 and eq4.

${x}^{2} - \frac{1}{2} x + \frac{1}{2} = 0$

or

$2 {x}^{2} - x + 1 = 0$

Thus,

The quadratic equation is $\bf 2 {x}^{2} - x + 1 = 0$.

Learn more:

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