Question

Can someone please help I really need these points

Answer 1

Given:

Two chords intersect each other inside the circle.

To find:

The value of x.

Solution:

According to intersecting chords theorem, if two chords intersect each other inside the circle, then the product of two segments of one chord is equal to  the product of two segments of second chord.

In the given circle,

$AE\times CE=BE\times DE$

$(3x-11)\times (5x-4)=(x+2)\times (-x+17)$

$15x^2-12x-55x+44=-x^2+17x-2x+34$

$15x^2-67x+44+x^2-15x-34=0$

$16x^2-82x+10=0$

Divide both sides by 2.

$8x^2-41x+5=0$

Splitting the middle term, we get

$8x^2-40x-x+5=0$

$8x(x-5)-1(x-5)=0$

$(8x-1)(x-5)=0$

Using zero product property, we get

$(8x-1)=0$ or $(x-5)=0$

$x=\dfrac{1}{8}$ or $x=5$

For $x=\dfrac{1}{8}$, the side AE is negative. So, $x=\dfrac{1}{8}$ is not possible.

Therefore, the required solution is $x=5$.