How do you solve #x^2+28x+196=0#?

3 Answers
Apr 28, 2018

Answer:

#x=-14" (repeated"#

Explanation:

#x^2+28x+196" is a "color(blue)"perfect square"#

#"note that "(a+b)^2=a^2+2ab+b^2#

#"with "a=x" and "b=14#

#rArrx^2+28x+196=(x+14)^2#

#"solve "(x+14)^2=0#

#rArrx+14=0rArrx=-14" (repeated)"#

Apr 28, 2018

Answer:

Either by inspection, graphing, or the quadratic formula.

Explanation:

Solve by Graphing

Every quadratic is a function on a graph. Just replace 0 with y= and graph the quadratic as a function. the x-coordinate(s) wherever the function intersects the x axis are your solution(s). in this instance, there's only 1 x-intercept and therefore only one solution.

Solve by Inspection

#x^2+28x+196#

#ax^2+bx+c#
#a=1,b=28,c=196#

Ask yourself, what factors of #c# multiplied by the factors of #a# both multiply together to make #c# and add to #b#. This sounds confusing, but when I lay it out, it will make more sense...

Think, if #sqrt196 = 14#, does #14+14 = 28#? Yes!

Then factor out the quadratic into 2 binomials...

#(x + 14)(x+14)#
.. and rearrange each individually to equal #x#

#x+14=0 => x = -14#

Quadratic formula

If the quadratic doesn't look easy to factor, then pull out the quadratic formula!

#ax^2+bx+c=0#
#a=1,b=28,c=196#

#x=(-b±sqrt(b^2-4ac))/(2a)#

#x=(-28±sqrt(28^2-4(28)(1)))/(2(1))#
#x=-14#

#±# is literally just a symbol for saying "plus or minus",

Hope it helps!

Apr 28, 2018

Answer:

The solution of this given quadratic equation is #(x+14)^2#

Explanation:

In the given expression we see the polynomials have the following numbers as their coefficients #1,28# and #196# so we can factorise this expression by middle term factorisation.
In middle term factorisation we have to split lower degree polynomial into two in our #28x# into the sum of two factors of #196# which when multiplied with each other give #196#
Here those two factors are #14# and #14#,as #14^2=196# and when you add #14# twice you get #28#.
#x^2+28x+196#
= #x^2+14x+14x+196#
= #x(x+14)+14(x+14)#
= #(x+14)(x+14)#
= #(x+14)^2#
I hope this was helpful. :)