Question #75911

4 Answers
Feb 11, 2018

Answer:

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#

Feb 11, 2018

Answer:

#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:

desmos.com/calculator

#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#.

Feb 11, 2018

Answer:

#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)#

Tony B

Feb 11, 2018

Answer:

#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#