Math, asked by Anonymous, 1 month ago

plz answer any 2questions from this
don't answer useless need correct
answer​

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Answered by dishamalhan325
25

Step-by-step explanation:

I knewed 12. and 16. hope it helps :-)

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Answered by nithyashree1577
2

Answer:

12. option a) equal roots

13.The equation is 4x^2+2x-1=0 let the roots are m&n.Given m=a and we have to prove n=4a^3-3a

4x^2+2x-1=0,or x^2+1/2x-1/4=0 now we can write m+n=a+n=-1/2, mn=an=-1/4. Now we try to find a-n to evaluate exact value of n & a. a-n=[(a+n)^2-4an]^1/2=[(-1/2)^2-4.(-1/4)]^1/2=(1/4+1)^1/2=(5)^1/2/2

4x^2+2x-1=0,or x^2+1/2x-1/4=0 now we can write m+n=a+n=-1/2, mn=an=-1/4. Now we try to find a-n to evaluate exact value of n & a. a-n=[(a+n)^2-4an]^1/2=[(-1/2)^2-4.(-1/4)]^1/2=(1/4+1)^1/2=(5)^1/2/2

14. Solution: Let D_1 andD_2 be the discriminants of the given equation.

x^2+px+q=0 and x^2+rx+s=0 , respectively,

Now, D_1+D_2=p^2-4q+r^2-4s

=p^2+r^2-4(q+s)

=p^2+r^2-2pr [ecause pr=2(p+s)]

=(p-r)^2geq 0Rightarrow D_1+D_2geq 0

Hence, at least one of the equations

x^2+px+q=0 and x^2+rx+s=0 has real root.

15.D1 = b2 - 4ac D2 = d2 + 4 ac ac is either +ve or negative so at least one of D1 & D2 is +ve so at least two real roots.

16. f(c) = - ve f(d) = + ve .'. There exists two real and distinct roots one in the interval (a, b) and other in (c, d). Hence, statement is true.

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