# Question 75911

Feb 11, 2018

Zeroes at $x = - 1 , x = \frac{5}{9}$

#### Explanation:

$\textcolor{b l u e}{\left(2 x - 5\right) \left(2 x - 5\right)} - 49 {x}^{2} = 0$

What I have in blue, we can multiply it using $\textcolor{b l u e}{F O I L} :$

• Multiply the first terms $\left(2 x \mathmr{and} 2 x\right)$ to get $4 {x}^{2}$

• The outside terms $\left(2 x \mathmr{and} - 5\right)$ to get $- 10 x$

• The inner terms $\left(- 5 \mathmr{and} 2 x\right)$ to get $- 10 x$

• The last terms $\left(- 5 \mathmr{and} - 5\right)$ to get $25$

$4 {x}^{2} - 10 x - 10 x + 25 - 49 {x}^{2} = 0$

Combining like terms, we get:

$- 45 {x}^{2} - 20 x + 25 = 0$

Now, at this point, there are a few things we could do, however the easiest may be to use the quadratic formula.

Quadratic Formula: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

When a quadratic equation is in standard form, it is of the form:
$a {x}^{2} + b x + c$, where a, b and c are all integers. Now, let's pattern match:

$a {x}^{2} + b x + c$
$- 45 {x}^{2} - 20 x + 25$ $\left(a = - 45 , b = - 20 , c = 25\right)$
Now, let's plug into the quadratic formula:

$\frac{20 \pm \sqrt{{\left(- 20\right)}^{2} - 4 \left(- 45\right) \left(25\right)}}{2 \left(- 45\right)}$

Let's simplify now:

$\frac{20 \pm \sqrt{400 + 4500}}{- 90}$

$\frac{20 \pm \sqrt{4900}}{-} 90$

$\frac{20 \pm 70}{-} 90$

All of these terms have a 10 in common, so we can divide each by 10.

$\frac{2 \pm 7}{-} 9$

We'll have 2 solutions:

1)$\frac{2 + 7}{-} 9 = \frac{9}{-} 9 = - 1$
$\textcolor{b l u e}{x = - 1}$ is one of our zeroes

2)$\frac{2 - 7}{-} 9 = \frac{- 5}{-} 9 = \frac{5}{9}$

$\textcolor{red}{\frac{5}{9}}$ is our second zero

Our two zeroes (solutions to this equation) are $x = - 1$ and $x = \frac{5}{9}$

Feb 11, 2018

$x = \frac{5}{9} , x = - 1$

#### Explanation:

using FOIL method to expand brackets:

${\left(2 x - 5\right)}^{2} = \left(2 x - 5\right) \left(2 x - 5\right)$

$= 4 {x}^{2} - 10 x - 10 x + 25$

$= 4 {x}^{2} - 20 x + 25$

collect like terms:

$4 {x}^{2} - 20 x + 25 - 49 {x}^{2} = 0$

$- 45 {x}^{2} - 20 x + 25 = 0$

$- 9 {x}^{2} - 4 x + 5 = 0$

factorise by grouping:

$- 9 {x}^{2} - 4 x + 5 = 0$

find two numbers that multiply to make the product of the first and last numbers, and add to make the middle number.

$- 9 + 5 = - 4$

$- 9 \cdot 5 = - 45$

make the second number (the coefficient of $x$) the sum of these two numbers.

$- 9 {x}^{2} - 9 x + 5 x + 5 = 0$

find common factors in each expression:

$- 9 {x}^{2} - 9 x = - 9 x \left(x + 1\right)$

$5 x + 5 = 5 \left(x + 1\right)$

add these together and simplify with a common factor:

$- 9 x \left(x + 1\right) + 5 \left(x + 1\right) = \left(- 9 x + 5\right) \left(x + 1\right)$

$\left(- 9 x + 5\right) \left(x + 1\right) = 0$

then solve for $x :$

$- 9 x + 5 = 0$ or $x + 1 = 0$

$- 9 x = - 5$ or $x = - 1$

$x = \frac{5}{9} , x = - 1$

this is the graph:

$y = 49 {x}^{2}$ is the graph in blue, and $y = {\left(2 x - 5\right)}^{2}$ is the graph in red.

the two graphs meet where $x = - 1$ and $x = 0.5555 . .$ or $\frac{5}{9}$.

Feb 11, 2018

${x}_{\text{intercepts")=-1 and +5/9 color(white)("ddd")color(brown)( larr " at } y = 0}$

#### Explanation:

We need to combine the ${x}^{2}$ terms. To do this we have to multiply out the brackets

$\textcolor{g r e e n}{{\left(2 x - 5\right)}^{2}} \textcolor{b r o w n}{- 49 {x}^{2}}$

$\textcolor{g r e e n}{4 {x}^{2} - 20 x + 25} \textcolor{b r o w n}{- {49}^{2}} = 0$

But $4 {x}^{2} - 49 {x}^{2} = - 45 {x}^{2}$ giving

$\textcolor{g r e e n}{- 45 {x}^{2} - 20 x + 25 = 0} \textcolor{red}{\leftarrow \text{corrected "+20x" to } - 20 x}$

$\textcolor{b r o w n}{\text{Note that as "-45x^2" is negative the graph is of form } \bigcap}$

Notice that 5 will divide exactly into all the values on the left of the equals sign. So divide all of both sides by 5.

Note that $\frac{0}{5}$ is still $0$

$\textcolor{g r e e n}{- 9 {x}^{2} - 4 x + 5 = 0}$

I chose to use the formula

$0 = y = a {x}^{2} + b x + c \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where a=-9;color(white)("d") b=-4 and c=+5

$x = \frac{+ 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(- 9\right) \left(+ 5\right)}}{2 \left(- 9\right)}$

$x = \frac{+ 4 \pm \sqrt{16 + 180}}{- 18}$

$x = - \frac{4}{18} \pm \frac{14}{18}$

$x = - \frac{2}{9} \pm \frac{7}{9}$

${x}_{\text{intercepts")=-1 and +5/9 color(white)("ddd}} \textcolor{b r o w n}{\leftarrow y = 0}$

Feb 11, 2018

$x = - 1 \text{ or } x = \frac{5}{9}$

#### Explanation:

${\left(2 x - 5\right)}^{2} - 49 {x}^{2} \text{ is a "color(blue)"difference of squares}$

•color(white)(x)a^2-b^2=(a-b)(a+b)#

$\text{here "a=2x-5" and } b = 7 x$

$\Rightarrow {\left(2 x - 5\right)}^{2} - 49 {x}^{2}$

$= \left(2 x - 5 - 7 x\right) \left(2 x - 5 + 7 x\right)$

$= \left(- 5 x - 5\right) \left(9 x - 5\right)$

$\Rightarrow \left(- 5 x - 5\right) \left(9 x - 5\right) = 0$

$\text{equate each factor to zero and solve for x}$

$- 5 x - 5 = 0 \Rightarrow x = - 1$

$9 x - 5 = 0 \Rightarrow x = \frac{5}{9}$