Q17. - Algebra 1

Question

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

$-x^2+6x-9=0$

Step-by-Step Solution

Answer

The root of the equation $-x^2+6x-9=0$ is $x=3$.

Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $f\left(x\right)=a{x}^2+bx+c$, where, $a\ne 0$ is called the standard form of the quadratic function.

Step 2. Rewrite the equation − x 2 + 6 x − 9 = 0 in the form f x = a x 2 + b x + c .

Write the equation $-x^2+6x-9=0$ in standard form.

This equation is already in standard form.

Write the equation $-x^2+6x-9=0$ in the form $f\left(x\right)=a{x}^2+bx+c$.

$f\left(x\right)=-x^2+6x-9$

Step 3. Plot the graph of the function f x = − x 2 + 6 x − 9 .

The graph of the function $f\left(x\right)=-x^2+6x-9$ is shown below.

Step 4. Solve the equation x 2 − 3 x − 4 = 0 from the graph of the function f x = x 2 − 3 x − 4 .

Observe the graph of the function $f\left(x\right)=x^2-3x-4$.

The graph intersects the $x$- axis exactly at one point $x=3$.  

Therefore the root of the equation $x^2-3x-4=0$ is $x=3$.