Question

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?

Answer 1

Answer:

0.897

Step-by-step explanation:

Brahmagupta formula for quadratic equation $ax^2+bx=c$ is

$x=\dfrac{\sqrt{4ac+b^2}-b}{2a}$

It involved only one solution.

The given equation is

$3x^2+4x=6$

Here, a=3, b=4 and c=6. Put these values in the above formula.

$x=\dfrac{\sqrt{4(3)(6)+(4)^2}-4}{2(3)}$

$x=\dfrac{\sqrt{4(3)(6)+(4)^2}-4}{2(3)}$

$x=\dfrac{\sqrt{72+16}-4}{6}$

$x=\dfrac{\sqrt{88}-4}{6}$

$x\approx \dfrac{5.38}{6}$

$x\approx 0.897$

Therefore, the required solution is 0.897.