Chapter 3.1, Problem 2E

To determine

The roots of the equation $a{x}^2+bx+c=0$ .

Answer to Problem 2E

The zeros, solution or the roots of a equation represent the same values, and the $y$ -intercept of the graph is $-10$ .

Explanation of Solution

Given information:

The given equation $a{x}^2+bx+c=0$ .

Calculation:

The given function is $f\left(x\right)=x^2+3x-10$ .

The zeros , solution and the roots or the y intercept of the equation represent the same values of the function.

Equate the given function with zero.

  $\begin{array}{c}{x}^2+3x-10=0\\ x=\frac{-3\pm \sqrt{3^2-4\times 1\times -10}}{2\times 1}\\ =\frac{-3\pm \sqrt{9+40}}{2}\\ =\frac{-3\pm \sqrt{49}}{2}\\ =\frac{-3\pm 7}{2}\\ =-5,2\end{array}$............... (1)

Find the zeros of $f\left(x\right)=x^2+3x-10$ .

The zeros of the function are the points at which the function value is zero.

Substitute $-5$ for $x$ in the above function.

  $\begin{array}{c}f\left(x\right)={\left(-5\right)}^2+3\left(-5\right)-10\\ =25-15-10\\ =0\end{array}$

Similarly for $x=2$ , $f\left(x\right)=0$ .

Now, find the solution of the equation $x^2+3x-10=0$ .

Similar to the zeros, here also the value of the function should be equal to zero.

Thus, the solutions of the equation are $-5\text{and}2$ .

Find the roots of the equation $10-x^2=3x$ .

  $\begin{array}{c}10-x^2=3x\\ x^2+3x-10=0\end{array}$

From (1), the roots of the equation are $-5\text{and}2$ .

Now, find the $y$ -intercept of the graph $y=\left(x+5\right)\left(x-2\right)$ .

For the $y$ -intercept, the value of $x$ must be zero.

Substitute 0 for $x$ in the equation $y=\left(x+5\right)\left(x-2\right)$ .

  $\begin{array}{c}y=\left(x+5\right)\left(x-2\right)\\ =5\times \left(-2\right)\\ =-10\end{array}$

Hence, the zeros, solution or the roots of a equation represent the same values, and the $y$ -intercept of the graph is $-10$ .