Question

the product of 2 consecutive even integers is 528 .represent the situation in the form of quadratic equation

Answer 1

Hey there!

We can first form an equation for this, as follows:

$\left(x\right)\left(x+2\right) = 528$

$x^{2} + 2x = 528$

$1x^{2} + 2x - 528 = 0$

Then, we can use the quadratic formula with $a = 1$, $b = 2$, $c = -528$ as follows:

$x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

$x = \dfrac{-2 \pm \sqrt{(-2)^{2} - 4(1)(-528)}}{2(1)}$

$x = \dfrac{-2 \pm \sqrt{4 - 4(-528)}}{2}$

$x = \dfrac{-2 \pm \sqrt{4(1 + 528))}}{2}$

$x = \dfrac{-2 \pm \sqrt{4(529)}}{2}$

$x = \dfrac{-2 \pm \sqrt{4(529)}}{2}$

$x = \dfrac{-2 \pm (2 \times 23)}{2}$

$x = -1 \pm (1 \times 23)$

$x = -1 \pm 23$

Now, we have two roots, which are:

$x = -1 + 23 = +22$

$x = -1 - 23 = -24$

We have to take the positive root of this. So we arrive with this:

$x = +22$

Now, apply this in the original equation:

$\left(x\right)\left(x+2\right) = 528$

$\left(22\right)\left(22+2\right) = 528$

$22 \times 24 = 528$

$528 = 528$

Hence, the equality is verified.

Answer 2

Step-by-step explanation:

Pls refer the given answer