Answer :
The nature of the roots of [tex]\mathbf{49x^2 - 28x + 4 = 0}[/tex] is (a) one real root
The equation is given as:
[tex]\mathbf{49x^2 - 28x + 4 = 0}[/tex]
To determine the nature of the roots, we make use of determinants.
The determinant (d) of a quadratic equation is calculated
[tex]\mathbf{d = b^2 - 4ac}[/tex]
Where the quadratic equation is:
[tex]\mathbf{ax^2 + bx + c = 0}[/tex]
By comparison, we have:
[tex]\mathbf{a = 49,\ b = -28,\ c = 4}[/tex]
So, we have:
[tex]\mathbf{d = b^2 - 4ac}[/tex]
[tex]\mathbf{d = (-28)^2 - 4 \times 49 \times 4}[/tex]
[tex]\mathbf{d = 784 - 784}[/tex]
[tex]\mathbf{d = 0}[/tex]
When d = 0, then the function has one real root.
Hence, the nature of the roots of [tex]\mathbf{49x^2 - 28x + 4 = 0}[/tex] is (a) one real root
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