Advertisements
Advertisements
प्रश्न
`x^2-4ax-b^2+4a^2=0`
उत्तर
We write, `-4ax=-(b+2a)x+(b-2 a)x as`
`x^2xx(-b^2+4a^2)=(-b^2+4a^2)x^2=-(b+2a)x xx(b-2a)x`
∴`x^2-4ax-b^2+4a^2=0`
⇒`x^2-(b+2a)x(b-2a)x-(b-2a)(b+2a)=0`
⇒`x[x-(b+2a)]+(b-2a)[x-(b+2a)]=0`
⇒`[x-(b+2a)] [x+(b-2a)]=0`
⇒`x-(b+2a)=0 or x+(b-2a)=0`
⇒`x=2a+b or x=-(b-2a) `
⇒`x=2a+b or x=2a-b`
Hence, (2a+b) and (2a-b) are the roots of the given equation.
APPEARS IN
संबंधित प्रश्न
Check whether the following is the quadratic equation:
x2 - 2x = (-2)(3 - x)
Solve `x/3 + 3/(6 - x) = (2(6 + x))/15; (x ≠ 6)`
Solve `((2x - 3)/(x -1)) - 4((x - 1)/(2x - 3)) = 3`
If x = − 3 and` x = 2/3 `are solution of quadratic equation `mx^2 + 7x + n = 0`, find the values of m and n.
` x^2+6x-(a^2+2a-8)=0`
`12abx^2-(9a^2-8b^2)x-6ab=0`
Solve:
`3sqrt(2x^2) - 5x - sqrt2 = 0`
Solve, using formula:
x2 + x – (a + 2)(a + 1) = 0
Write the following quadratic equation in standard form ax2 + bx + c = 0 : x (x + 3) = 7
The value (values) of x satisfying the equation x2 – 6x – 16 = 0 is ______.
