Question

Prove that the equation 2(a²+b²)x² + 2(a+b)x + 1 = 0 has no real root if a≠b​

Answer 1

Step-by-step explanation:

for real roots,

b²-4ac>=0

(2(a+b))²-4(2(a²+b²))>=0

4a²+4b²+8ab-8a²-8b²>=0

-(4a²+4b²-8ab)>=0

-(2a-2b)²>=0

(2a-2b)²<=0

the minimum value of any perfect square is 0 it is never negative for real numbers.

therefore only possible case,

2a-2b=0

2a=2b

a=b

therefore 2(a²+b²)x² + 2(a+b)x + 1 = 0 has no real root if a≠b

Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm \: 2( {a}^{2} + {b}^{2})x + 2(a + b)x + 1 = 0$

So, on comparing with quadratic equation Ax² + Bx + C = 0, we get

$\rm \: A = 2( {a}^{2} + {b}^{2})$

$\rm \: B \: = \: 2(a + b)$

$\rm \: C \: = \: 1$

Now, to check the nature of roots of quadratic equation, we have to check the sign of Discriminant.

So, Consider

$\rm \: Discriminant, \: D$

$\rm \: = \: {B}^{2} - 4AC$

$\rm \: = \: [2(a + b)]^{2} - 4 \times 1 \times 2( {a}^{2} + {b}^{2})$

$\rm \: = \: 4(a + b)^{2} - 8( {a}^{2} + {b}^{2})$

$\rm \: = \: 4[(a + b)^{2} - 2( {a}^{2} + {b}^{2})]$

$\rm \: = \: 4[ {a}^{2} + {b}^{2} + 2ab - 2{a}^{2} - 2{b}^{2}]$

$\rm \: = \: 4[ - {a}^{2} - {b}^{2} + 2ab ]$

$\rm \: = \: - 4[ {a}^{2} + {b}^{2} - 2ab ]$

$\rm \: = \: - 4[ {(a - b)}^{2} ]$

$\rm \: < 0 \: \: if \: a \: \ne \: b$

$\rm\implies \:D < 0 \: \: if \: a \: \ne \: b$

$\rm\implies \:2( {a}^{2} + {b}^{2}) {x}^{2} + 2(a + b)x + 1 = 0 \: have \: no \: real \: roots \: if \: a \: \ne \: b$

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Concept Used :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac