Question

Find the number of solutions of $5^x=x^2+x+1$.

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Answer 1

Here, Let's consider F(x) = $5^x$
g(x) = $x^2+x+1$
Let's draw the graph of F(x) = $5^x$

We can relate this graph from F(x) = $a^x$ where $a>1$

g(x) = $x^2+x+1$

Draw the graph of $x^2+x+1$

Draw the quadratic equations,we need to know when the graph cut s x -axis

And coordinate of vertex of graph

$x^2+x+1$

Discriminant of $x^2+x+1$

$D=b^2-4ac=1-4=-3$ (No real solution)

since discriminant is negative $x^2+x+1$ will have no real solution.

vertex of graph $x^2+x+1$ is

x-coordinate=$\frac{-b}{2a}=\frac{-1}{2}$

y-coordinate=$\frac{-D}{4a}=\frac{-(-3)}{4\times 1}=\frac{3}{4}$

and at $x=0,y=0+0+1=1$

at $x=1,y=1+1+1=3$

Graph $x^2+x+1$ is

we can see that at x=0

Both the graph cuts at 1

so,there will be only 1 solution for $5^x=x^2+x+1$