Question

The quadratic equations ${x^2} + 15|x| + 14 = 0$ has
1)only positive solutions
2)only negative solutions
3)no solution
4)both positive and negative solutions

Answer 1

We have to find the solution of the given quadratic equation${x^2} + 15|x| + 14 = 0$. We solve this question using the concept of real roots of an equation . We will firstly put $x > 0$and then find the roots of the quadratic equation and then put $x < 0$and then find the roots of the quadratic equations . Solving the equation for the two cases gives the real roots of the solution .

Complete step-by-step solution:
Given : ${x^2} + 15|x| + 14 = 0$
Let us consider two cases such as $x > 0$and$x < 0$. As removing mod we add $\pm$ after removing the mod function .
Case 1 ) For , $x > 0$
${x^2} + 15x + 14 = 0$
We find the roots of the quadratic equation using the formula of quadratic formula .
I.e.
$x = \dfrac{{[ - b \pm \sqrt {{b^2} - 4ac} ]}}{{2a}}$
Where a is the coefficient of ${x^2},b$is the coefficient of $x$ and $c$ is the coefficient of the constant term .
Using the above formula , we get the value of $x$ as
$x = \dfrac{{[ - 15 \pm \sqrt {({{( - 15)}^2} - 4 \times 1 \times 14)} ]}}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{[ - 15 \pm \sqrt {(225 - 56)} ]}}{2}$
$\Rightarrow x = \dfrac{{[ - 15 \pm \sqrt {(169)} ]}}{2}$
$\Rightarrow x = \dfrac{{\left[ { - 15 \pm 13} \right]}}{2}$
$\Rightarrow x = \dfrac{{\left[ { - 15 + 13} \right]}}{2}$or $x = \dfrac{{\left[ { - 15 - 13} \right]}}{2}$
$\Rightarrow x = - 1orx = - 14$
As for solving the quadratic equations for$x > 0$, we don’t get any real value of $x$ .
Hence , there is no solution for$x > 0$.
Case 2 ) For , $x < 0$
${x^2} - 15x + 14 = 0$
We find the roots of the quadratic equation using the formula of quadratic formula .
I.e.
$x = \dfrac{{[ - b \pm \sqrt {({b^2} - 4ac)} ]}}{{2a}}$
Where a is the coefficient of ${x^2},b$is the coefficient of $x$ and $c$ is the coefficient of the constant term .
Using the above formula , we get the value of $x$ as
$x = \dfrac{15\pm{\sqrt {({{(15)}^2} - 4 \times 1 \times 14)} }}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{[15 \pm \sqrt {(225 - 56)} ]}}{2}$
$\Rightarrow x = \dfrac{{[15 \pm \sqrt {(169)} ]}}{2}$
$\Rightarrow x = \dfrac{{\left[ {15 \pm 13} \right]}}{2}$
$\Rightarrow x = \dfrac{{\left[ {15 + 13} \right]}}{2}or x = \dfrac{{\left[ {15 - 13} \right]}}{2}$
$\Rightarrow x = 1orx = 14$
As for solving the quadratic equations for $x < 0$, we don’t get any real value of $x$ .
Hence , there is no solution for $x < 0$.
From the two cases we conclude that there is no solution for the quadratic equation .
Thus , the equation ${x^2} + 15|x| + 14 = 0$ has no solution .
Hence , the correct option is $\left( 3 \right)$.

Note: Since ${b^2} - 4ac$ determines whether the quadratic equation $a{x^2} + bx + c = 0$ has real roots , ${b^2} - 4ac$ is called the discriminant of this quadratic equation .
(1) two distinct real roots , if ${b^2} - 4ac > 0$
(2) two equal real roots , if ${b^2} - 4ac = 0$
(3) no real roots , if ${b^2} - 4ac < 0$