How do you solve #5x^2+19x=4#?

Question

Eva Abramson · Accepted Answer

#x = 1/5 or x = -4#

As it is a quadratic equation # rArr # make = 0

#5x^2 + 19x -4 = 0#

Find factors of 5 and 4 which subtract to make 19. Signs will be different - more positive. Cross multiply and subtract

#color(white)(xxx)(5)" "(4)#

#color(white)(xxxx)color(magenta)(5)" "color(lime)(1 rarr 1xx1 =1)# #color(white)(xxxx)color(lime)(1)" "color(magenta)(4 rarr 5xx4=20)" "20-1 = 19#

We have the right factors - now fill in the signs.

#color(white)(xxxx)color(red)(5" "-1) rarr 1xx-1 =-1# #color(white)(xxxx)color(blue)(1" "+4) rarr 5xx+4=+20" "+20-1 = +19#

#color(red)((5x-1))color(blue)((x+4))=0#

Either factor could be equal to 0.

If #color(red)(5x-1 = 0, rarr 5x =1 rarr x = 1/5#

If #color(blue)( x+4=0 rarr x = -4#

Genesis Allen · Answer

undefined To solve the equation 5x^2 + 19x = 4, follow these steps:

1. Rewrite the equation in standard form: 5x^2 + 19x - 4 = 0.
2. Use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 5, b = 19, and c = -4.
3. Substitute the values of a, b, and c into the quadratic formula.
4. Calculate the discriminant, Δ = b^2 - 4ac.
5. If the discriminant is positive, there are two real solutions. If it's zero, there is one real solution (a repeated root). If it's negative, there are no real solutions (two complex roots).
6. Plug the values of a, b, c, and Δ into the quadratic formula and simplify to find the solutions for x.