Chapter 1.3, Problem 7QR

To determine

To Find: solution of the equation $2x^2+9x-5=0$ using factoring

Answer to Problem 7QR

Solution of $2x^2+9x-5=0$ are $x=-5,\frac{1}{2}$

Explanation of Solution

Given:

The equation, $2x^2+9x-5=0$

Concept used:

For any quadratic equation $a{x}^2+bx+c=0$ , where $a$ and $b$ are coefficient and $c$ is constant provided, $a\ne 0$

Then the equation is solved by splitting the middle term such that both the term that sum together to obtain $b$ multiplied together to obtain the product equal to the product of $a$ and $c$ .

Calculation:

In order to factorize the given equation $2x^2+9x-5=0$ ,

Split the middle term such that each term that sum together to obtain $9$ multiplied together to obtain the product equal to the product of $2$ and $-5$ .

  $\begin{array}{c}2x^2+9x-5=0\\ 2x^2+10x-x-5=0\mathrm{...}(i)\end{array}$

Such That $10x+\left(-x\right)=9x$

And $10x\times \left(-x\right)=2x^2\times \left(-5\right)$

Simplify the equation $\left(i\right)$ to solve the given equation:

  $\begin{array}{c}2x^2+10x-x-5=0\\ 2x\left(x+5\right)-\left(x+5\right)=0\text{factor}\text{out}\text{the}\text{commom}\text{term}\\ \left(x+5\right)\left(2x-1\right)=0\text{factor}\text{out}\text{common}\text{binomial}\text{parantheses}\end{array}$

Since product of two terms is zero.

Hence, either of the term or both of term are zero

Thus,

  $\begin{array}{l}x+5=0\text{or}2x-1=0\\ x=-5\text{or}x=\frac{1}{2}\end{array}$

Thus, the solution of $2x^2+9x-5=0$ are $x=-5,\frac{1}{2}$

Conclusion:

Solution of $2x^2+9x-5=0$ are $x=-5,\frac{1}{2}$