Question

Find the value of $k$ so that the quadratic equation ${x^2} + 2(k + 1)x + k + 5 = 0$ has at least one positive root?

Answer 1

Here we have to find the value of $k$ so that the quadratic equation ${x^2} + 2(k + 1)x + k + 5 = 0$ has at least one positive root. So, we will use a quadratic formula to solve this equation. Quadratic formula states that if $a{x^2} + bx + c = 0$ is a quadratic equation, so the root of this equation is given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $a$ and $b$ are the coefficients of ${x^2}$ and $x$ respectively and $c$ is a constant. For a quadratic equation to have at least one positive root the discriminant i.e., ${b^2} - 4ac$ must be equal to zero for one positive root and should be greater than zero for two positive roots.

Complete step by step answer:
In this question we have asked to find the value of $k$ so that the quadratic equation ${x^2} + 2(k + 1)x + k + 5 = 0$ has at least one positive root.
The given equation is a quadratic equation. Quadratic equations are those equations that consist of at least one term which is squared and also called a second degree equation. The general form of a quadratic equation is $a{x^2} + bx + c = 0$ where $a,b$ and $c$ are numerical constants or coefficients, and $x$ is an unknown variable.
Comparing the given equation by general form of a quadratic equation $a{x^2} + bx + c = 0$.
We get, $a = 1,\,\,b = 2(k + 1) = 2k + 2$ and $c = k + 5$
For a quadratic equation to have at least one positive root the discriminant must be equal to zero for one positive root and should be greater than zero for two positive roots.
From the quadratic formula, we have the discriminant ${b^2} - 4ac$.
So, we need ${b^2} \geqslant 4ac$
So, in the given equation we have,
$\Rightarrow {(2k + 2)^2} \geqslant (4 \times 1 \times (k + 5))$
$\Rightarrow 4{k^2} + 4k + 4 \geqslant 4k + 20$
Simplifying the above equation by dividing the whole equation by $4$. We get,
$\Rightarrow {k^2} + k + 1 \geqslant k + 5$
Cancelling the $k$ terms. We get,
$\Rightarrow {k^2} + 1 \geqslant 5$
$\Rightarrow {k^2} \geqslant 4$
So, if $k = 4$ then
$\Rightarrow {x^2} + 2(4 + 1)x + (4 + 5) = 0$
Simplifying the above equation. We get,
$\Rightarrow {x^2} + 8x + 2x + 9 = 0$
Adding $x$ terms. We get,
$\Rightarrow {x^2} + 10x + 9 = 0$
Solve the above equation by the method of factorization. we get,
$\Rightarrow {x^2} + 9x + x + 9 = 0$
Taking $x$ and $1$ as a common factor. We get
$\Rightarrow x(x + 9) + 1(x + 9)$
$\Rightarrow (x + 9)(x + 1) = 0$
Therefore, $x = - 9$ and $x = - 1$
These are both negative values. So, we cannot take the value $k = 4$.
It seems like we have to cancel the $c$ part. So, if $k = - 5$ then the equation would be
$\Rightarrow {x^2} + 2( - 5 + 1)x + ( - 5 + 5) = 0$
$\Rightarrow {x^2} - 10x - 2x = 0$
Simplifying the $x$ terms. We get,
$\Rightarrow {x^2} - 8x = 0$
Solve the above equation by the method of factorization. We get,
$\Rightarrow x(x - 8) = 0$
Therefore, we get $x = 0$ and $x = 8$
As $8$ is a positive root so, $k = - 5$ must be the solution of the given quadratic equation.

Note: Note: Note that we only use quadratic formulas to solve the quadratic equation if we can’t factorize the equation by quadratic formula. If in the problem we have asked if the equation has equal and real roots then the discriminant i.e., ${b^2} - 4ac$ of a quadratic equation must be equal to zero. Always remember that the number of roots of a quadratic equation depends on the degree of the equation.