Question

pls solve it immediately

Answer 1

1) given,

x²+px-3=0 { a=1, b=p, c=-3 }

let the roots of the polynomial be r and s

r+s = -b/a

rs = c/a

sum of the squares of roots = 10

=>r²+s² = 10

=>(r+s)²-2rs = 10

=>(-b/a)² - 2(c/a) = 10

=>b²/a² -2c/a = 10

=>p²/1² - 2(-3)/1 = 10

=>p²+6 = 10

=>p²=4

=>p=±2 (a)

3) given,

roots for incorrect equation are r = 3 and s = 2

then the equation is x²-(r+s)x+rs=0

=>x²-(3+2)x+3(2)=0

=>x²-5x+6=0

The correct equation is x²-5x-6=0

=>x²-6x+x-6=0

=>x(x-6)+1(x-6)=0

=>(x-6)(x+1)=0

=>x=6 or x=-1 (d)

4) I think it has no solution (a)

HOPE U UNDERSTAND

PLS MARK IT AS BRAINLIEST​

Answer 2

1)
Given quadratic polymomial,
x^+px-3=0
Let 'a' and 'b' be the zeroes,
we have,
a+b=-p...........................................1
ab=-3..............................................2
As per given condition we have,
a^2+b^2=10
(a+b)^2-2ab=10
By putting eq1 and eq2 we have,
p^2+6=10
p^2=4
p=±2
Hence value of p=±2





2)
Given quadratic polynomial,
a(b-c)x^2+b(c-a)x+c(a-b)=0

(ab-ac)x^2+(bc-ab)x+(ac-bc)=0

(ab-ac)x^2+{-(ab-ac)-(ac-bc)}x+(ac-bc)=0

(ab-ac)x^2-(ab-ac)x-(ac-bc)x+(ac-bc)=0

(ab-ac)x{x-1}-(ac-bc){x-1}=0

{(ab-ac)x-(ac-bc)}{x-1}=0

x=(ac-bc)/(ab-ac) (or) x=1

x=c(a-b)/a(b-c) (or) x=1

Hence roots of given quadratic equation are
{c(a-b)/a(b-c)} and 1.



3)
Roots of the quadratic equation which was formed by mistake are 3 and 2.
We know that quadratic equation is in the form of ,
=k{x^2-(sum of the zeroes)x+(product of zeroes)}

Quadratic equation,
x^2-5x+6=0

As per given condition we have,
Correct equation ,
x^2-5x-6=0
x^2-6x+x-6=0
x(x-6)+1(x-6)=0
(x+1)(x-6)=0
x=-1 or 6
Hence roots (or) zeroes are -1 and 6



4)
Given equation,
√(x+1)-√(x-1)=√(4x-1)
Squaring on both sides we get,
{√(x+1)}^2+{(x-1)}^2-2√(x^2-1)=4x-1

2x-2√(x^2-1)=4x-1

-2√(x^2-1)=2x-1

√(x^2-1)=(1-2x)/2

x^2-1=(1+4x^2-4x)/4

4x^2-4=1+4x^2-4x

-5=-4x

x=5/4

Hence the given equation has only one solution.