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# Solve For X (X-1)/(X-2)+(X-3)/(X-4)=3 1/3; X ≠ 2, 4 - CBSE Class 10 - Mathematics

ConceptRelationship Between Discriminant and Nature of Roots

#### Question

Solve for x

(x-1)/(x-2)+(x-3)/(x-4)=3 1/3; x ≠ 2, 4

#### Solution

We have been given,

(x-1)/(x-2)+(x-3)/(x-4)=3 1/3; x ≠ 2, 4

Now we solve the above equation as follows,

((x-1)(x-4)+(x-3)(x-2))/((x-2)(x-4))=10/3

(x^2-5x+4+x^2-5x+6)/(x^2-6x+8)=10/3

6x2 - 30x + 30 = 10x2 - 60x + 80

4x2 - 30x + 50 = 0

2x2 - 15x + 25 = 0

Now we also know that for an equation ax2 + bx + c = 0, the discriminant is given by the following equation:

D = b2 - 4ac

Now, according to the equation given to us, we have,a = 2, b = -15 and c = 25.

Therefore, the discriminant is given as,

D = (-15)2 - 4(2)(25)

= 225 - 200

= 25

Now, the roots of an equation is given by the following equation,

x=(-b+-sqrtD)/(2a)

Therefore, the roots of the equation are given as follows,

x=(-(-15)+-sqrt25)/(2(2))

=(15+-5)/4

Now we solve both cases for the two values of x. So, we have,

x=(15+5)/4

= 5

Also,

x=(15-5)/4

=5/2

Therefore, the value of x = 5, 5/2

Is there an error in this question or solution?

#### APPEARS IN

Solution Solve For X (X-1)/(X-2)+(X-3)/(X-4)=3 1/3; X ≠ 2, 4 Concept: Relationship Between Discriminant and Nature of Roots.
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