Answer
425.1k+ views
Hint: Before attempting this question, one should have prior knowledge about the quadratic equation as in the above equation is a quadratic equation of degree 2 so this equation can easily be solved by the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and also remember to let ${p^2}$ as the perfect square of the given equation and use the algebraic identities ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$.
Complete step-by-step solution:
The given quadratic equation is: ${x^2} + ax + a + 1 = 0$
So, by applying quadratic formula we get
$x = \dfrac{{ - a \pm \sqrt {{a^2} - 4(a + 1)} }}{{2a}}$
= $x = \dfrac{{ - a \pm \sqrt {{a^2} - 4a - 4} }}{{2a}}$
So, to get the integral values of a the expression ${a^2} - 4a - 4$ need to be a perfect square
So ${a^2} - 4a - 4 = {p^2}$
Here ${p^2}$ is a perfect square
\[
{a^2} - 4a - 4 - {p^2} = 0 \\
{a^2} - 4a = 4 + {p^2} \\
{a^2} - 4a + 4 = 4 + 4 + {p^2} \\
\]
Here 4 is added to both left- and right-hand sides
As we know that ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$
Therefore ${(a - 2)^2} = 8 + {p^2}$
Let the value of p=1
$
{(a - 2)^2} = 8 + 1 \\
{(a - 2)^2} = 9 \\
a - 2 = \pm 3 \\
$
Let a – 2 = 3
This gives a = 5
And a – 2 = -3
This gives a = - 1
So, the options A and D are correct
We can further put the ${p^2} = 4,9,16..$ and other perfect squares but the answer will be irrational which is irreverent for us.
Note: In the above question to approach the solution we used the quadratic formula i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where a, b, and c is the real number of an quadratic equation than to get the integral roots of a the determinant ($\sqrt {{b^2} - 4ac} $) must to be a perfect square so we put $\sqrt {{b^2} - 4ac} = {p^2}$ where ${p^2}$ is a perfect square.
Complete step-by-step solution:
The given quadratic equation is: ${x^2} + ax + a + 1 = 0$
So, by applying quadratic formula we get
$x = \dfrac{{ - a \pm \sqrt {{a^2} - 4(a + 1)} }}{{2a}}$
= $x = \dfrac{{ - a \pm \sqrt {{a^2} - 4a - 4} }}{{2a}}$
So, to get the integral values of a the expression ${a^2} - 4a - 4$ need to be a perfect square
So ${a^2} - 4a - 4 = {p^2}$
Here ${p^2}$ is a perfect square
\[
{a^2} - 4a - 4 - {p^2} = 0 \\
{a^2} - 4a = 4 + {p^2} \\
{a^2} - 4a + 4 = 4 + 4 + {p^2} \\
\]
Here 4 is added to both left- and right-hand sides
As we know that ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$
Therefore ${(a - 2)^2} = 8 + {p^2}$
Let the value of p=1
$
{(a - 2)^2} = 8 + 1 \\
{(a - 2)^2} = 9 \\
a - 2 = \pm 3 \\
$
Let a – 2 = 3
This gives a = 5
And a – 2 = -3
This gives a = - 1
So, the options A and D are correct
We can further put the ${p^2} = 4,9,16..$ and other perfect squares but the answer will be irrational which is irreverent for us.
Note: In the above question to approach the solution we used the quadratic formula i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where a, b, and c is the real number of an quadratic equation than to get the integral roots of a the determinant ($\sqrt {{b^2} - 4ac} $) must to be a perfect square so we put $\sqrt {{b^2} - 4ac} = {p^2}$ where ${p^2}$ is a perfect square.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)