How do you solve quadratic equation #4x^2+11x-20=0#?

Question

Eva Adamson · Accepted Answer

#x = 5/4 and x = 4#

The first method to check for solving a quadratic equation is whether the expression factorises.

#4x^2 +11x-20 = 0#

"Find factors of 4 and 20 which subtract to make 11"

Note that 11 is ODD, so the factors must combine to give one ODD and and one even number.

That immediately eliminates #2xx2# and #10xx2# as possible factors of 4 and 20 ( because their multiples will always be even.)

When trying different combinations, remember not to have a common factor in any horizontal row.

Find factors and cross-multiply. Subtract the products to get 11.

#" "ul(4" "20)# #" "4" "5" "rarr1 xx 5 = 5# #" "1" "4" "rarr 4xx4 = ul16# #color(white)(xxxxxxxxxxxxxxxxxxx)11# the difference is 11

We have the correct factors, now work with the signs.

MINUS #20# means that the signs must be different. PLUS #11# means there must be more positives. Fill in the correct signs, starting from #color(red)(+11)#

#color(red)(+11) " "rarr color(red)(+16) and color(blue)(-5)#

#" "ul(4" "20)# #" "4" "5" "rarr 1 xx 5 = color(blue)(-5)# #" "1" "4" "rarr 4xx4 = ulcolor(red)(+16)# #color(white)(xxxxxxxxxxxxxxxxxxx)color(red)(+11)#

Now fill in the signs next to the correct factors:

#" "ul(4" "20)# #" "4" "color(blue)(-5)" "rarr 1 xx color(blue)(-5) = color(blue)(-5)# #" "1" "color(red)(+4)" "rarr 4xxcolor(red)(+4) = ulcolor(red)(+16)# #color(white)(xxxxxxxxxxxxxxxxxxxxx)color(red)(+11)#

Now you have the factors: Top row is one bracket and bottom row is the other factor.

#4x^2 +11x-20 = 0# #(4x-5)(x+4) =0#

Letting each factor be equal to 0 gives the 2 solutions

#4x-5 =0 " "rarr " " 4x=5 rarr x= 5/4# #x+4=0" "rarr" "x=-4#

Avery Alexander · Answer

#5/4 and -4# #y = 4x^2 + 11x - 20 = 0#. Use the new Transforming Method (Socratic Search) Transformed equation: #y' = x^2 + 11x - 80 = 0#. First, find the 2 real roots of y' that have opposite signs (ac < 0). Then, divide them by (a). Factor pairs of (- 80) --> ...(4, - 20)(5, -16). This last sum is (-11 = -b). There for the 2 real roots of y' are: 5 and - 16. The 2 real roots of y are: #5/4# and #- 16/4 = - 4#

Genesis Allen · Answer

undefined To solve the quadratic equation $4x^2 + 11x - 20 = 0$, you can use the quadratic formula, which is $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

For the given equation, $a = 4$, $b = 11$, and $c = -20$. Substituting these values into the quadratic formula, you get:

$$x = \frac{{-11 \pm \sqrt{{11^2 - 4(4)(-20)}}}}{{2(4)}}$$

Simplify the expression under the square root:

$$11^2 - 4(4)(-20) = 121 + 320 = 441$$

Substitute back into the equation:

$$x = \frac{{-11 \pm \sqrt{{441}}}}{{8}}$$

$$x = \frac{{-11 \pm 21}}{{8}}$$

So the solutions are:

$$x_1 = \frac{{-11 + 21}}{{8}} = \frac{{10}}{{8}} = \frac{{5}}{{4}}$$

$$x_2 = \frac{{-11 - 21}}{{8}} = \frac{{-32}}{{8}} = -4$$

Therefore, the solutions to the quadratic equation are $x = \frac{5}{4}$ and $x = -4$.