For the first equation, x² - 4x + 3√2 = 0, we identify a=1, b=-4, and c=3√2. We then calculate the discriminant D. Since 12√2 ≈ 12 × 1.414 = 16.968, D = 16 - 16.968 = -0.968, which is less than 0. Therefore, this equation does not have real roots.

For the first equation, x² - 4x + 3√2 = 0, we identify a=1, b=-4, and c=3√2. We then calculate the discriminant D. Since 12√2 ≈ 12 × 1.414 = 16.968, D = 16 - 16.968 = -0.968, which is less than 0. Therefore, this equation does not have real roots.

For the second equation, x² + 4x - 3√2 = 0, we have a=1, b=4, and c=-3√2. Calculating the discriminant, we get D = 16 + 12√2. Since 12√2 is a positive number, D will be greater than 0, meaning this equation has real roots.

For the second equation, x² + 4x - 3√2 = 0, we have a=1, b=4, and c=-3√2. Calculating the discriminant, we get D = 16 + 12√2. Since 12√2 is a positive number, D will be greater than 0, meaning this equation has real roots.

For the third equation, x² - 4x - 3√2 = 0, we have a=1, b=-4, and c=-3√2. The discriminant is D = 16 + 12√2. This is also a positive value, indicating that this equation has real roots.

For the fourth equation, 3x² + 4√3x + 4 = 0, we have a=3, b=4√3, and c=4. The discriminant is D = (4√3)² - 4(3)(4) = 48 - 48 = 0. Since D=0, this equation has real and equal roots.

Which one of the following equations does not have real roots ?

Answer: The equation $x^2 - 4x + 3\sqrt{2} = 0$ does not have real roots.

Step-by-step solution

Step 1: Understand the condition for real roots

A quadratic equation of the form $ax^2 + bx + c = 0$ has real roots if its discriminant, denoted by $D$ , is greater than or equal to zero. The discriminant is calculated using the formula $D = b^2 - 4ac$ . If $D < 0$ , the roots are not real; they are complex.

Step 2: Analyze the first equation

For the first equation, $x^2 - 4x + 3\sqrt{2} = 0$ , we identify $a=1$ , $b=-4$ , and $c=3\sqrt{2}$ . We then calculate the discriminant $D$ . Since $12\sqrt{2} \approx 12 \times 1.414 = 16.968$ , $D = 16 - 16.968 = -0.968$ , which is less than 0. Therefore, this equation does not have real roots.

Step 3: Analyze the second equation

For the second equation, $x^2 + 4x - 3\sqrt{2} = 0$ , we have $a=1$ , $b=4$ , and $c=-3\sqrt{2}$ . Calculating the discriminant, we get $D = 16 + 12\sqrt{2}$ . Since $12\sqrt{2}$ is a positive number, $D$ will be greater than 0, meaning this equation has real roots.

Step 4: Analyze the third equation

For the third equation, $x^2 - 4x - 3\sqrt{2} = 0$ , we have $a=1$ , $b=-4$ , and $c=-3\sqrt{2}$ . The discriminant is $D = 16 + 12\sqrt{2}$ . This is also a positive value, indicating that this equation has real roots.

Step 5: Analyze the fourth equation

For the fourth equation, $3x^2 + 4\sqrt{3}x + 4 = 0$ , we have $a=3$ , $b=4\sqrt{3}$ , and $c=4$ . The discriminant is $D = (4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$ . Since $D=0$ , this equation has real and equal roots.

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