Advertisements
Advertisements
Question
Solve the following equation by factorisation :
`sqrt(3x^2 - 2x - 1) = 2x - 2`
Advertisements
Solution
`sqrt(3x^2 - 2x - 1) = 2x - 2`
Squaring both sides
3x2 - 2x – 1 = (2x - 2)2
⇒ 3x2 – 2x – 1 = 4x2 – 8x + 4
⇒ 4x2 – 8x + 4 – 3x2 + 2x + 1 = 0
⇒ x2 – 6x + 5 = 0
⇒ x2 – 5x – x + 5 = 0
⇒ x(x – 5) –1(x – 5) = 0
⇒ (x – 5)(x – 1) = 0
Either x – 5 = 0,
then x = 5
or
x – 1 = 0,
then x = 1
Check :
(i) If x = 5, then
L.H.S. = `sqrt(3x^2 - 2x - 1)`
= `sqrt(3 xx (5)^2 - 2 xx 5 - 1)`
= `sqrt(3 xx 25 - 10 - 1)`
= `sqrt(75 - 10 - 1)`
= `sqrt(64)`
= 8
R.H.S. = 2x – 2
= 2 x 5 – 2
= 10 – 2
= 8
∵ L.H.S. = R.H.S.
∴ x = 5 is a root
(ii) If x = 1, then
L.H.S. = `sqrt(3x^2 - 2x - 1)`
= `sqrt(3(1)^2 - 2(1) - 1)`
= `sqrt(3 xx 1 - 2 - 1)`
= `sqrt(3 - 2 - 1)`
= 0
R.H.S. = 2x – 2
= 2 x 1 – 2
= 2 – 2
= 0
∵ L.H.S. = R.H.S.
∴ x = 1 is also its root
Hence x = 5, 1.
APPEARS IN
RELATED QUESTIONS
`2x^2+5x-3=0`
Solve the following quadratic equation by factorisation.
x2 + x – 20 = 0
Solve the following quadratic equations by factorization:
\[\frac{1}{x - 3} + \frac{2}{x - 2} = \frac{8}{x}; x \neq 0, 2, 3\]
Find the values of k for which the roots are real and equal in the following equation:
\[4x^2 + kx + 3 = 0\]
If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax2 +bx + 1 = 0 having real roots is
If one root the equation 2x2 + kx + 4 = 0 is 2, then the other root is
Solve the following equation by factorization
3x2 – 5x – 12 = 0
Solve the following equation by factorization
`(1)/(x + 6) + (1)/(x - 10) = (3)/(x - 4)`
2x articles cost Rs. (5x + 54) and (x + 2) similar articles cost Rs. (10x – 4), find x.
Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.
