How do you use the quadratic formula to find the solutions of #x^2-21=4x# ?

3 Answers
Dec 31, 2017

#x = 7" "# or #" "x = -3#

Explanation:

Given:

#x^2-21=4x#

Subtract #4x# from both sides to get:

#x^2-4x-21 = 0#

This is now in standard form:

#ax^2+bx+c = 0#

with #a=1#, #b=-4# and #c=-21#.

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#

Dec 31, 2017

#x=-3" or "x=7#

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#

Dec 31, 2017

See below.

Explanation:

First you need to make sure if the equation is in the form of #ax^2 + bx + c = 0# where #a != 0#.

#x^2 - 21 = 4x#
#rArr x^2 - 4x - 21 = 0#

Now it is in the desired form. Let's compare this equation with the general form.

After comparing, we get,

#a = 1#, #b = -4#, #c = -21#

Now Let's find out the Discriminant.

#D# = #b^2 - 4ac# = #(-4)^2 - 4 * 1 * (-21)# = #16 + 84# = #100 > 0#

That means, the equation has two real, and distinct roots.

Now Applying Quadratic Formula (Sridhar Acharya's Formula) :

#alpha = (- b + sqrt(D))/(2a) = (4 + sqrt(100))/(2) = 7#

#beta = (-b - sqrt(D))/(2a) = (4 - sqrt(100))/2 = -3#

These are the two roots, #7# and# -3#.

Hence Explained.