Question

The function f(x) = 2x2 + 3x + 5, when evaluated, gives a value of 19. What is the function’s input value?

Answer 1

$f(x)=\quad 2{ x }^{ 2 }+3x+5$

Let's called the input 'z'

When we plug 'z' in the function we get ;

$f(z)=\quad 2{ z }^{ 2 }+3z+5$

And we know that, this is equal to 19, so ;

$2{ z }^{ 2 }+3z+5=\quad 19$

Let's rearrange the equation.

$2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0$

So we have a quadratic equation here.

We'll use this formula to solve it :

$\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

The formula is used in equation formed like this :

$a{ x }^{ 2 }+bx+c=0$

In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

$a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }$

So,

$z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }$

So the input can be both, 2 and $-\frac{7}{2}$