How do you solve #x^2+28x+196=0#?
#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#
Either by inspection, graphing, or the quadratic formula.
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
Ask yourself, what factors of
Then factor out the quadratic into 2 binomials...
.. and rearrange each individually to equal
If the quadratic doesn't look easy to factor, then pull out the quadratic formula!
Hope it helps!
The solution of this given quadratic equation is
In the given expression we see the polynomials have the following numbers as their coefficients
In middle term factorisation we have to split lower degree polynomial into two in our
Here those two factors are
I hope this was helpful. :)