Question

Which of these Quadratic equation do not have real roots?
a) x^2-4x+3root2=0

b) x^2+4x-3root2=0

c) x^2-4x-3root2=0

d) 3x^2+4root3x+4=0
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Answer 1

Step-by-step explanation:

x² - 4x + 3 = 0

x² -3x -1x + 3 = 0

x(x-3) - 1 (x-3) = 0

(x-3) (x-1) = 0

x - 3 = 0 or x - 1 = 0

x = 3 or x = 1

Answer 2

$\big\langle\Big\langle\bigg\langle\Bigg\langle\LARGE\underbrace{\underline{\sf{Understanding\;the\;Question}}}\big\rangle\Big\rangle\bigg\rangle\Bigg\rangle$

Here is the concept that a quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable. One absolute rule is that the first constant "a" cannot be a zero.

Let's do it !!

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★ Correct Question :-

Prove that which of these Quadratic equation do not have real roots,

a) x^2-4x+3root2=0

b) x^2+4x-3root2=0

c) x^2-4x-3root2=0

d) 3x^2+4root3x+4=0

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★ Formula To Complete:-

b2-4ac=0

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★Solution-

The quadratic equation $ax^{2}+bx+c=0$ has no real roots only when discriminant D<0

D=$b^{2} - 4ac<0$

A) $x^{2} +4x+\sqrt[3]{2} = 0$

$Here, D=(4) ^{2} -4\times\sqrt[3]{2}=16-16.97<0$

$B) x^{2} +4x-\sqrt[3]{2} =0$

Here,

$D=(4) ^{2} -4\times(-\sqrt[3]{2} )=16+16.97>0$

$C) x^{2} +5x+\sqrt[3]{2} =0$

Here,  

$D=(5)^{2} -4\times(\sqrt[3]{2})=25-16.97>0$

$D) 3x^{2} +\sqrt[4]{3x} +4=0$

Here,  

$D=(\sqrt[4]{3}^{2} ) 2 -4\times3\times4=48-48=0$

Hence, option A.