Question

(2x+3)(2x+5)(x-1)(x-2)=30 the roots available are​

Answer 1

Answer:

if LHS=RHS you should,

2x+3+2x+5+x-1+x-2=30

4x+8+x-1+x-2=30

6x+8-3=30

6x+5=30

6x=30-5

6x=25

x=25/6

x=4

so substitute 4 in all x and find value

Answer 2

Answer:

$\large\boxed{X=0,X=\frac{1}{2},X=\frac{-1+\sqrt{105} }{4},X=\frac{-1-\sqrt{105} }{4}    }$

Step-by-step explanation:

$\textnormal{To find the value of x, first, you have to isolate it on one side of the equation.}$

First, expand the form to solve with distributive property.

$\textnormal{Distributive property=\boxed{a(b+c)=ab+ac}}$.

$\displaystyle (2x+3)(2x+5)(x-1)(x-2)=4x^2+4x^3-25x^2-13x+30$

Next, rewrite the problem down.

$\displaystyle 4x^4+4x^3-25x^2-13x+30=30$

Then, subtract 30 from both sides.

$\displaystyle 4x^4+4x^3-25x^2-13x+30-30=30-30$

Solve.

$\displaystyle 30-30=0$

$\displaystyle 4x^4+4x^3-25x^2-13x=0$

You can also solve by the factors.

$\displaystyle 4x^4+4x^3-25x^2-13x=x(2x+1)(2x^2+x-13)$

$\displaystyle 4x^4+4x^3-25x^2-13x$

Factor it out by the common term of x.

$\displaystyle x(4x^3+4x^2-25x-13)$

$\displaystyle4x^3+4x^2-25x-13=(2x+1)(2x^2+x-13)$

$\displaystyle(2x+1)2x^2+x-13=0$

Solve.

$\displaystyle 2x+1=0$

Subtract 1 from both sides.

$\displaystyle 2x+1-1=0-1$

Solve.

$\displaystyle 2x=-1$

Divide by 2 from both sides.

$\displaystyle \frac{2x}{2}=\frac{-1}{2}$

Solve.

$\displaystyle x=-\frac{1}{2}$

Solve with $\displaystyle2x^2+x-13=0$.

$\displaystyle 2x^2+x-13=0=\boxed{X=\frac{-1+\sqrt{105} }{4}, X=\frac{-1-\sqrt{105} }{4}  }$

$\boxed{\textnormal{Therefore, the correct answer is \boxed{X=0, X=-\frac{1}{2}, X=\frac{-1+\sqrt{105} }{4}, X=\frac{-1-\sqrt{105} }{4}   }}}.$