# The nature of the roots of the equation \[{{x}^{2}}-5x+7=0\] is

(a) No real roots

(b) 1 real root and 1 imaginary

(c) Real and unequal

(d) Real and equal

Last updated date: 25th Mar 2023

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Answer

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Hint: First of all, compare the given equation with a general quadratic equation \[a{{x}^{2}}+bx+c\] and find the values of a, b and c. Then find the discriminant, \[D={{b}^{2}}-4ac\]. If D > 0, then roots are real and distinct. If D = 0, then roots are real and equal, and if D < 0, then roots are imaginary.

Complete step-by-step answer:

Here we have to find the nature of the roots of the equation, \[{{x}^{2}}-5x+7=0\].

Before, proceeding with the question, we must know what the roots, nature of roots, etc are. First of all, roots or zeros

of any equation are the values of the variable which satisfy the equation. In other words, if by substituting some value of the variable, the equation becomes zero, then that value of the variable is the root of the equation. For example, equation 5x + 10 = 0 has a root – 2 because by substituting x = – 2, this equation becomes 0.

The nature of roots tells us whether the roots of the equation are real or imaginary, distinct or coincident, etc.

For any general quadratic equation, \[a{{x}^{2}}+bx+c=0\] where \[a,b,c\in R\] and \[a\ne 0\], then, \[D={{b}^{2}}-4ac\] is known as the discriminant of the quadratic equation.

(i) For D > 0, roots are real and distinct (unequal)

(ii) For D = 0, roots are real and coincident (equal)

(iii), For D < 0, roots are imaginary

Let us consider the quadratic equation given in the question \[{{x}^{2}}-5x+7=0\]

By comparing it to the general quadratic equation \[a{{x}^{2}}+bx+c=0\], we get, a = 1, b = -5 and c = 7

We know that \[D={{b}^{2}}-4ac\].

By substituting the values of a, b and c, we get

\[D={{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 7 \right)\]

By simplifying RHS of the above equation, we get,

D = 25 – 28

D = – 3

Here, - 3 < 0 that means D < 0.

Hence, we get that the roots of the quadratic equation \[{{x}^{2}}-5x+7=0\] are imaginary in nature.

Note:

Some students make the mistake of writing discriminant D as \[\sqrt{{{b}^{2}}-4ac}\] while D is equal to \[{{b}^{2}}-4ac\]. Also, students should properly take the values of a, b, and c and especially take care of their signs while comparing the given quadratic equation with the general quadratic equation \[a{{x}^{2}}+bx+c\]. Here, \[a,b,c\in R\] and \[a\ne 0\].

Complete step-by-step answer:

Here we have to find the nature of the roots of the equation, \[{{x}^{2}}-5x+7=0\].

Before, proceeding with the question, we must know what the roots, nature of roots, etc are. First of all, roots or zeros

of any equation are the values of the variable which satisfy the equation. In other words, if by substituting some value of the variable, the equation becomes zero, then that value of the variable is the root of the equation. For example, equation 5x + 10 = 0 has a root – 2 because by substituting x = – 2, this equation becomes 0.

The nature of roots tells us whether the roots of the equation are real or imaginary, distinct or coincident, etc.

For any general quadratic equation, \[a{{x}^{2}}+bx+c=0\] where \[a,b,c\in R\] and \[a\ne 0\], then, \[D={{b}^{2}}-4ac\] is known as the discriminant of the quadratic equation.

(i) For D > 0, roots are real and distinct (unequal)

(ii) For D = 0, roots are real and coincident (equal)

(iii), For D < 0, roots are imaginary

Let us consider the quadratic equation given in the question \[{{x}^{2}}-5x+7=0\]

By comparing it to the general quadratic equation \[a{{x}^{2}}+bx+c=0\], we get, a = 1, b = -5 and c = 7

We know that \[D={{b}^{2}}-4ac\].

By substituting the values of a, b and c, we get

\[D={{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 7 \right)\]

By simplifying RHS of the above equation, we get,

D = 25 – 28

D = – 3

Here, - 3 < 0 that means D < 0.

Hence, we get that the roots of the quadratic equation \[{{x}^{2}}-5x+7=0\] are imaginary in nature.

Note:

Some students make the mistake of writing discriminant D as \[\sqrt{{{b}^{2}}-4ac}\] while D is equal to \[{{b}^{2}}-4ac\]. Also, students should properly take the values of a, b, and c and especially take care of their signs while comparing the given quadratic equation with the general quadratic equation \[a{{x}^{2}}+bx+c\]. Here, \[a,b,c\in R\] and \[a\ne 0\].

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