How do you use the quadratic formula to find the solutions of #x^2-21=4x# ?
3 Answers
Explanation:
Given:
#x^2-21=4x#
Subtract
#x^2-4x-21 = 0#
This is now in standard form:
#ax^2+bx+c = 0#
with
The roots are given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(1))(color(blue)(-21))))/(2(color(blue)(1)))#
#color(white)(x) = (4+-sqrt(16+84))/2#
#color(white)(x) = (4+-sqrt(100))/2#
#color(white)(x) = (4+-10)/2#
#color(white)(x) = 2+-5#
That is:
#x = 7" "# or#" "x = -3#
Explanation:
#"rearrange into standard form":ax^2+bx+c=0#
#rArrx^2-4x-21=0larrcolor(blue)"in standard form"#
#"with the equation in standard form we can solve using the"#
#color(blue)"quadratic formula"#
#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#
#"here "a=1,b=-4" and "c=-21#
#x=(4+-sqrt(16+84))/2#
#color(white)(x)=(4+-sqrt100)/2=(4+-10)/2=(-6)/2" or "14/2#
#rArrx=-3" or "x=7#
#"this can also be solved by factorising"#
#x^2-4x-21=0#
#"the factors of - 21 which sum to - 4 are - 7 and + 3"#
#rArr(x-7)(x+3)=0#
#"equate each factor to zero and solve for x"#
#x-7=0rArrx=7#
#x+3=0rArrx=-3#
See below.
Explanation:
First you need to make sure if the equation is in the form of
Now it is in the desired form. Let's compare this equation with the general form.
After comparing, we get,
Now Let's find out the Discriminant.
That means, the equation has two real, and distinct roots.
Now Applying Quadratic Formula (Sridhar Acharya's Formula) :
These are the two roots,
Hence Explained.