Answer:
1.686
Step-by-step explanation:
Given is an quadratic equation as
[tex]2x^2-x-4=0[/tex]
Also we are given the formula for general equation
[tex]ax^2+bx+c[/tex] as
[tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
Here in our equation we have
[tex]a=2\\b=-1\\c=-4[/tex]
Hence solution would be
[tex]\frac{1+\sqrt{1+32} }{4} ,\frac{1-\sqrt{1+32} }{4} \\\frac{1+\sqrt{33} }{4} ,\frac{1-\sqrt{33} }{4}[/tex]
Greater would be the one with positive square root
[tex]\frac{1+\sqrt{33} }{4} \\=1.6861\\=1.686[/tex]
Answer:
Brahmagupta’s solution to a quadratic equation of the form ax2 + bx = c involved only one solution. Which solution would he have found for the equation 3x2 + 4x = 6?
first option.
To find the solutions to an equation of the form ax2+bx+c , use the quadratic equation: x=−b±b2−4ac√2a . What are the two solutions to 2x2−x−4=0 ? Use the quadratic equation. (Report the greater of the two solutions first, using 3 significant digits for both solutions)
third option.
