To calculate: The solution of the equation 2(x3+1)2+5(x3+1)−12=0.

Question

George Burge · Accepted Answer

Calculation:  Consider the equation, 2(x3+1)2+5(x3+1)−12=0.  Now, simplify the equation 2(x3+1)2+5(x3+1)−12=0.  2(x3+1)2+5(x3+1)−12=0.  2{(x3)2+(1)2+2(x3)(1)}+5(x3+1)−12=0.  2{(x29)+1+2x3}+5x3+5−12=0.  2x29+9x3−5=0  Now, taking the least common multiple of the denominator,  2x29+9x3−5=0  2x2+27x−459=0  2x2+27x−45=0  This is a quadratic equation so compare the equation from the standard form of quadratic equation ax2+bx−c=0 and identify the values of a, b and c.  Here, a=2, b=27 and c=-45  Now, apply the quadratic formula,  x=−(27)±(27)2−4(2)(−45)2(2)  =−27±729+3604  =−27±10894  =−27±334  Firstly, consider the positive sign,  x=−27+334  =64  =32 Now, consider the negative sign,  x=−27−334  =−604  =-15  Now, to check the solution put the values of x in the original equation,  Substitute x=-15 in the equation 2(x3+1)2+5(x3+1)−12=0

To calculate: The solution of the equation 2(\frac{x}{3}+1)^{2}+5(\

Answered question

Answer & Explanation

New Questions in Pre-Algebra