Question

Question #75911

Answer 1

Zeroes at `x=-1, x=5/9`

Explanation:

`color (blue)((2x-5)(2x-5))-49x^2=0`

What I have in blue, we can multiply it using `color(blue)(FOIL):`

• Multiply the first terms `(2x and 2x)` to get `4x^2`

• The outside terms `(2x and -5)` to get `-10x`

• The inner terms `(-5 and 2x)` to get `-10x`

• The last terms `(-5 and -5)` to get `25`

`4x^2-10x-10x+25-49x^2=0`

Combining like terms, we get:

`-45x^2-20x+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: `(-b+- sqrt(b^2-4ac))/(2a)`

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

`ax^2+bx+c`
`-45x^2-20x+25` `(a= -45, b= -20, c= 25)`
Now, let's plug into the quadratic formula:

`(20+-sqrt((-20)^2-4(-45)(25)))/(2(-45))`

Let's simplify now:

`(20+-sqrt(400+4500))/(-90)`

Adding the terms in the radical:

`(20+-sqrt(4900))/-90`

Further simplifying the radical:

`(20+-70)/-90`

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

`(2+-7)/-9`

We'll have 2 solutions:

1)`(2+7)/-9= 9/-9=-1`
`color(blue)(x=-1)` is one of our zeroes

2)`(2-7)/-9= (-5)/-9= 5/9`

`color(red)(5/9)` is our second zero

Our two zeroes (solutions to this equation) are `x=-1` and `x=5/9`

Answer 2

`x = 5/9, x = -1`

Explanation:

using FOIL method to expand brackets:

`(2x-5)^2 = (2x-5)(2x-5)`

`= 4x^2 - 10x - 10x + 25`

`= 4x^2 - 20x + 25`

collect like terms:

`4x^2-20x + 25 - 49x^2 = 0`

`-45x^2 - 20x + 25 = 0`

`-9x^2 - 4x + 5 = 0`

factorise by grouping:

`-9x^2 - 4x + 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 * 5 = -45`

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

`-9x^2 - 9x + 5x + 5 = 0`

find common factors in each expression:

`-9x^2-9x = -9x(x+1)`

`5x + 5 = 5(x+1)`

add these together and simplify with a common factor:

`-9x(x+1) + 5(x+1) = (-9x+5)(x+1)`

`(-9x+5)(x+1) = 0`

then solve for `x:`

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

`-9x = -5` or `x = -1`

`x = 5/9, x = -1`

this is the graph:

`y = 49x^2` is the graph in blue, and `y = (2x-5)^2` is the graph in red.

the two graphs meet where `x = -1` and `x = 0.5555..` or `5/9`.

Answer 3

`x_("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

`color(green)((2x-5)^2)color(brown)(-49x^2)`

`color(green)(4x^2-20x+25)color(brown)(-49^2)=0`

But `4x^2-49x^2=-45x^2` giving

`color(green)(-45x^2-20x+25=0) color(red)(larr"corrected "+20x" to "-20x)`

`color(brown)("Note that as "-45x^2" is negative the graph is of form " nnn)`

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 `0/5` is still `0`

`color(green)(-9x^2-4x+5=0)`

I chose to use the formula

`0=y=ax^2+bx+c color(white)("dddd")->color(white)("dddd") x=(-b+-sqrt(b^2-4ac))/(2a)`

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

`x=(+4+-sqrt((-4)^2-4(-9)(+5)))/(2(-9))`

`x=(+4+-sqrt(16+180))/(-18)`

`x=-4/18+-14/18`

`x=-2/9+-7/9`

`x_("intercepts")=-1 and +5/9 color(white)("ddd")color(brown)( larr y=0)`

Answer 4

`x=-1" or "x=5/9`

Explanation:

`(2x-5)^2-49x^2" is a "color(blue)"difference of squares"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`"here "a=2x-5" and "b=7x`

`rArr(2x-5)^2-49x^2`

`=(2x-5-7x)(2x-5+7x)`

`=(-5x-5)(9x-5)`

`rArr(-5x-5)(9x-5)=0`

`"equate each factor to zero and solve for x"`

`-5x-5=0rArrx=-1`

`9x-5=0rArrx=5/9`