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27 May 2021, 08:00
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Difficulty:
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56% (01:33) correct
44%
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Which of the following equations has two different roots?
A. x^2−x+4=0
B. x^2−4x+4=0
C. x^2−3x+9=0
D. x^2−5x+5=0
E. x^2+3x+10=0
A. x^2−x+4=0
B. x^2−4x+4=0
C. x^2−3x+9=0
D. x^2−5x+5=0
E. x^2+3x+10=0
27 May 2021, 14:18
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Ans is C
To have 2 different real roots, the discriminant (b^2-4ac) >0, which is only satisfied in option C.
To have 2 different real roots, the discriminant (b^2-4ac) >0, which is only satisfied in option C.
29 May 2021, 04:48
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For a quadratic equation ax^2 + bx + c = 0 to have two different real roots, the discriminant, b^2 – 4ac > 0.
Now, let us analyze the nature of the roots of each of the quadratic equations given in the options.
Option A: x^2 – x + 4 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -1 and c = 4
Thus, b^2 – 4ac = (-1)^2 – 4 (1) (4) = -15 < 0
Option B: x^2 – 4x + 4 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -4 and c = 4
Thus, b^2 – 4ac = (-4)^2 – 4 (1) (4) = 0
Option C: x^2 – 3x + 9 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -3 and c = 9
Thus, b^2 – 4ac = (-3)^2 – 4 (1) (9) = -27 < 0
Option D: x^2 – 5x + 5 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -5 and c = 5
Thus, b^2 – 4ac = (-5)^2 – 4 (1) (5) = 5 > 0
Option E: x^2 + 3x + 10 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = 3 and c = 10
Thus, b^2 – 4ac = (3)^2 – 4 (1) (10) = -31 < 0
Clearly, Option D gives b^2 - 4ac > 0. Hence, the correct answer is Option D.
Now, let us analyze the nature of the roots of each of the quadratic equations given in the options.
Option A: x^2 – x + 4 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -1 and c = 4
Thus, b^2 – 4ac = (-1)^2 – 4 (1) (4) = -15 < 0
Option B: x^2 – 4x + 4 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -4 and c = 4
Thus, b^2 – 4ac = (-4)^2 – 4 (1) (4) = 0
Option C: x^2 – 3x + 9 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -3 and c = 9
Thus, b^2 – 4ac = (-3)^2 – 4 (1) (9) = -27 < 0
Option D: x^2 – 5x + 5 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = -5 and c = 5
Thus, b^2 – 4ac = (-5)^2 – 4 (1) (5) = 5 > 0
Option E: x^2 + 3x + 10 = 0
Comparing the above equation with the general form ax^2 + bx + c = 0, we get a = 1, b = 3 and c = 10
Thus, b^2 – 4ac = (3)^2 – 4 (1) (10) = -31 < 0
Clearly, Option D gives b^2 - 4ac > 0. Hence, the correct answer is Option D.
