Advertisements
Advertisements
Question
Solve the following equation by factorization
a2x2 + (a2+ b2)x + b2 = 0, a ≠ 0
Advertisements
Solution
a2x2 + (a2+ b2)x + b2 = 0
⇒ a2x(x + 1) + b²(x + 1) = 0
⇒ (x + 1) (a2x + b2) = 0.
⇒ (x + 1) = 0, then x = -1
or
a2x + b2 = 0, then a2x = - b2.
⇒ x = `(-b^2)/a^2`
Hence x = -1, `(-b^2)/a^2`.
APPEARS IN
RELATED QUESTIONS
Solve the equation `4/x-3=5/(2x+3); xne0,-3/2` for x .
A two-digit number is such that the products of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number?
Divide 57 into two parts whose product is 680.
Find the values of k for which the roots are real and equal in each of the following equation:
\[4 x^2 + px + 3 = 0\]
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
Solve the following : `("x" - 1/2)^2 = 4`
Find two consecutive natural numbers such that the sum of their squares is 61.
The sum of the numerator and denominator of a certain positive fraction is 8. If 2 is added to both the numerator and denominator, the fraction is increased by `(4)/(35)`. Find the fraction.
Solve the following equation by factorisation :
`sqrt(3)x^2 + 10x + 7sqrt(3)` = 0
Find the roots of the following quadratic equation by the factorisation method:
`2/5x^2 - x - 3/5 = 0`
