How do you solve 2x^2 + 6x - 5 = 0 ? I divided by 2 to get x^2 + 3x - 2.5, but I can't find two numbers that add to equal 3 and multiply to equal -2.5
Nope
Nobody said anything about a formula unless you mean "Take half he coefficient of x and square it and add it to both sides"
Do you do that before or after dividing by 2?
+2498 \(\text{let s use } (a+b)^2=a^2+2ab+b^2\text{ and }a^2-b^2=(a-b)(a+b):\\x^2 + 3x - 2.5=0\\x^2 + 3x - 2.5+4.75=0+4.75\\x^2 + 3x + 2.25=4.75\text{ (1.5*1.5=2.25 1.5*2=3x)}\\(x+1.5)^2=4.75\\(x+1.5)^2-4.75=0\\(x+1.5-\sqrt{4.75})*(x+1.5+\sqrt{4.75})=0\\x_1=0.679449\\x_2=-3.67945\)
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+33659 Divide by 2 first and take constant to the RHS: \(x^2+3x=\frac{5}{2}\)
Now add square of half the coefficient of x to both sides: \(x^2+3x+\frac{9}{4}=\frac{5}{2}+\frac{9}{4}\)
Rewrite as: \((x+\frac{3}{2})^2=\frac{19}{4}\)
You should be able to take it from here.
FOLLOW THE PROCEDURE OUTLINED IN THIS SOLUTION:
Solve for x:
2 x^2+6 x-5 = 0
Divide both sides by 2:
x^2+3 x-5/2 = 0
Add 5/2 to both sides:
x^2+3 x = 5/2
Add 9/4 to both sides:
x^2+3 x+9/4 = 19/4
Write the left hand side as a square:
(x+3/2)^2 = 19/4
Take the square root of both sides:
x+3/2 = sqrt(19)/2 or x+3/2 = -sqrt(19)/2
Subtract 3/2 from both sides:
x = sqrt(19)/2-3/2 or x+3/2 = -sqrt(19)/2
Subtract 3/2 from both sides:
Answer: |x = sqrt(19)/2-3/2 or x = -3/2-sqrt(19)/2
