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# Integral value of a that quadratic equation ${x^2} + ax + a + 1 = 0$ has integral roots is equal to(This question has multiple correct options)A). -1B). 2C). 1D). 5

Last updated date: 25th Jun 2024
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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.