Question

How do you solve `4x^2 +4x = 15` using the quadratic formula?

Answer 1

`x = 3/2, 5/2`.

Explanation:

First Of All, Convert the Equation to It's General Form `ax^2 + bx + c = 0`.

So We have,

`color(white)(xxx)4x^2 + 4x = 15`

`rArr 4x^2 + 4x - 15 = 0` [Subtract `15` from both sides.]

So, Comparing the Equation with the General Form, We get,

`a = 4, b = 4, c = -15`.

So, Let's Find the Discriminat.

`D = b^2 - 4ac = 4^2 - 4*4*(-15) = 16 + 240 = 256`

As `D gt 0`, we will get two roots which are real and distinct.

Now Use The Quadratic Formula or Sridhar Acharya's Rule (whatever you may call it in your country).

`alpha = (-b + sqrt(D))/(2a) = (-4 + sqrt(256))/(2 * 4) = (-4 + 16)/(8) = 3/2`

And `beta = (-b - sqrt(D))/(2a) = (-4 - sqrt(256))/(2 * 4) = (-4 - 16)/(8) = 5/2`

So, `x = 3/2, 5/2`

Hope this helps.

Answer 2

`3/2 " or" - 5/2`

Explanation:

Make the expression equal to zero:

`4 x^2 + 4x - 15 = 0`

The quadratic formula is:

`x = (-b +- sqrt (b^2 -4 a c )) / (2a)`

In our case we substitute:

`a = 4, b = 4, c = -15`

So that the quadratic formula becomes:

`x = (-4 +- sqrt (4^2 -4 * 4 *(-15) )) / (2*4)`

`= (-4 +- sqrt (16 + 240 )) / (8)`

`= (-4 +- 16) / (8)`

`= 3/2 "or" - 5/2`

We can also solve this by factorising (here you need to guess)

`4 x^2 + 4x - 15 = 0`

`(2x + 5 ) (2x - 3) = 0`

Then make both parenthesis equal to zero to get the same answers:

`(2x + 5 ) = 0 " or " (2x - 3) = 0`