Question

How do you solve this type of problem, what would be the procedure? Assume PQRS is a parallelogram, if PQ=x2−10 and SR=3x , find all possible values of x.

Answer 1

Answer:

$x = 5$

Step-by-step explanation:

Given

Shape: Parallelogram PQRS

$PQ = x^2 - 10$

$SR = 3x$

Required

Find all possible values of x

Every parallelogram have parallel and equal opposite sides;

From the given parameters, one can easily deduce that PQ and SR are opposite sides;

This implies that

$PQ = SR$

Substitute values of PQ and SR

$x^2 - 10 = 3x$

Subtract 3x from both sides

$x^2 - 10 -3x = 3x - 3x$

$x^2 - 10 -3x = 0$

Rearrange

$x^2 -3x- 10 = 0$

Now, we have a quadratic equation;

Expand the above expression

$x^2 +2x-5x- 10 = 0$

Factorize

$x(x+2)-5(x+2) = 0$

$(x-5)(x+2) = 0$

$x - 5 = 0$   or    $x + 2 = 0$

Solve for x in both cases

$x = 5$ or $x = -2$

But x can't be negative;

So, the possible value of x is

$x = 5$