Advertisements
Advertisements
प्रश्न
If a and b are distinct real numbers, show that the quadratic equations
`2(a^2+b^2)x^2+2(a+b)x+1=0` has no real roots.
उत्तर
The given equation is `2(a^2+b^2)x^2+(a+b)x+1=0`
∴`D=[2(a+b)]^2-4xx2(a^2+b^2)xx1`
=`4(a^+2ab+b^2)-8(a^+b^2)`
=`4a^2+8ab+4b^2-8a^2-8b^2`
=`-4a^2+8ab-4b^2`
=`-4(a^2-2ab+b^2)`
=`-4(a-b)^2<0`
Hence, the given equation has no real roots.
APPEARS IN
संबंधित प्रश्न
Solve for x
`(x-1)/(x-2)+(x-3)/(x-4)=3 1/3`; x ≠ 2, 4
Find the value of k for which x = 1is a root of the equation `x^2+kx+3=0`
`(2x-3) (3x+1)=0`
`x-1/x=3,x≠0`
`x^2-2ax-(4b^2-a^2)=0`
If 3 is a root of the quadratic equation` x^2-x+k=0` find the value of p so that the roots of the equation `x^2+2kx+(k^2+2k+p)=0` are equal.
Find the value of p for which the quadratic equation `2x^2+px+8=0` has real roots.
If the roots of the equation `(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0`are equal, prove that `a/b=c/d`
For what values of k, the roots of the quadratic equation (k + 4) x2 + (k + 1) x + 1 = 0 are equal ?
Solve for x: \[\frac{1}{x - 3} - \frac{1}{x + 5} = \frac{1}{6}, x \neq 3, - 5\]