From 5x^2-6x+8, subtract 3x^2-2x-4?

Mathematics

1 answer:

victus00 [196]3 years ago

8 0

$(5x^2-6x+8)-(3x^2-2x-4)=\\5x^2-6x+8-3x^2+2x+4=2x^2-4x+12$

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X^2 + 32x + 256 = 1<br>please show steps

Elena L [17]

$\boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}$

<h2>Explanation:</h2>

In this case, we have the following equation:

$x^2+32x+256=1$

But we can write this equation as:

$x^2+32x+256=1 \\ \\ Subtract \ -1 \ from\ both \ sides: \\ \\ x^2+32x+256-1=1-1 \\ \\ x^2+32x+255=0$

So this final result is a quadratic equation written in Standard Form ( $ax^2+bx+c=0$ ). We need to find the solutions to this equations, so let's use quadratic formula:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ b=32 \\ c=255 \\ \\ \\ x=\frac{-32 \pm \sqrt{(32)^2-4(1)(255)}}{2(1)} \\ \\ x=\frac{-32 \pm \sqrt{1024-1020}}{2} \\ \\ x=\frac{-32 \pm \sqrt{4}}{2} \\ \\ x=\frac{-32 \pm 2}{2} \\ \\ Finally, \ two \ solutions: \\ \\ \boxed{x_{1}=-15} \\ \\ \boxed{x_{2}=-17}$