Question

\(x^2\) - 3\(x^2\) + 4 = 0

2\(x^2\) + 5\(x\) - 7 = 0

8\(x\) - 4 = 0

4 - 5\(x\) = 0

3\(x^4\) + 7\(x^2\) - 10 = 0

Answer 1

a: \(x^2-3x^2+4=0\)

=>\(4-2x^2=0\)

=>\(2-x^2=0\)

=>\(x^2=2\)

=>\(x=\pm\sqrt{2}\)

b: \(2x^2+5x-7=0\)

=>\(2x^2+7x-2x-7=0\)

=>(2x+7)(x-1)=0

=>\(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=1\end{matrix}\right.\)

c: \(8x-4=0\)

=>8x=4

=>\(x=\dfrac{4}{8}=\dfrac{1}{2}\)

d: 4-5x=0

=>5x=4

=>\(x=\dfrac{4}{5}\)

e: \(3x^4+7x^2-10=0\)

=>\(3x^4+10x^2-3x^2-10=0\)

=>\(\left(3x^2+10\right)\left(x^2-1\right)=0\)

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(x\in\left\{1;-1\right\}\)