Question

find the roots 2x²+5x-12=0

Answer 1

Find the roots 2x²+5x-12=0  your ans is-  x=3/2 and x=-8

Answer 2

Answer:

Required roots of the given equation are $\frac{ - 5 + \sqrt{73} }{2}$

or $\frac{ - 5 - \sqrt{73} }{2}$

Step-by-step explanation:

Given,

${x}^{2} + 5x - 12 = 0$

We can solve this by Sridhar Acharya's formula.

We know by Sridhar Acharya's formula,

if $a {x}^{2} + bx + c = 0$ is a equation then $x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$b = 5 \\ a = 1 \\ c = - 12$

So,

$x = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4 \times 1 \times ( - 12)} }{2 \times 1} \\ = \frac{ - 5 \pm \sqrt{25 + 48} }{2} \\ = \frac{ - 5 \pm \sqrt{73} }{2}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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