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Write whether the following statements are true or false. Justify your answers.

(i) Every quadratic equation has exactly one root.

(ii) Every quadratic equation has at least one real root.

(iii) Every quadratic equation has at least two roots.

(iv) Every quadratic equations has at most two roots.

(v) If the coefficient of x² and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

(vi)If the coefficient of x² and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

Answers (2)

(i) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation $x^{2}-4=0$
$\\x^{2}-4=0\\ x^{2}=4\\ x=\pm \sqrt{4}\\ x=2 \: \: \: \: \: \: \: x=-2$

Here – 2, 2 are the two roots of the equation.

Hence it is false that every quadratic equation has exactly one root.

(ii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation

$x^{2}-x+2=0$
$x^{2}-x+2=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

$\\a=1,b=-1,c=2\\ b^{2}-4ac=(-1)^{2}-4(1)(2)\\ =1-8\\ =-7\\ b^{2}-4ac<0$

Hence both the roots of the equation are imaginary.

Hence the statement every quadratic equation has at least one real root is False.

(iii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation $x^{2}-4x+4=0$

$\\x^{2}-4x+4=0\\ x^{2}+(2)^{2}-2(x)(2)\\ (x-2)^{2}=0\: \: \: \: \: \: \: \: \: \: \: \: \: \: (using\: (a-b)^{2}=a^{2}+b^{2}-2ab)\\ (x-2)=0\\ x=2$

The equation $x^{2}-4x+4=0$ has only one root which is x = 2

Hence the given statement is False

(iv) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

As we know that the standard form of a quadratic equation is $ax^{2} + bx + c = 0$

It is a polynomial of degree 2.

As per the power of x is 2. There is at most 2 values of x exist that satisfy the equation.

Hence the given statement every quadratic equation has at most two roots is true.

(v) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + x - 2 = 0$

$x^{2} + x - 2 = 0$ is a quadratic equation in which the coefficient of x² and constant term have opposite signs.

$x^{2} + x - 2 = 0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$\\a=1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(1)(-2)\\ =1+8\\ =9\\ b^{2}-4ac>0$

Here $b^{2}-4ac>0$

Hence these types of equations have real roots.

(vi) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + 4= 0$

In $x^{2} + 4= 0$ , the coefficient of x² and the constant term has the same sign and coefficient of x is 0.

$x^{2} + 4= 0$

Compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$\\a=1,b=0,c=4\\ b^{2}-4ac=(0)^{2}-4(1)(4)\\ =16\\ b^{2}-4ac<0$

Here $b^{2}-4ac<0$

Hence these types of equations have no real roots.

Posted by

infoexpert24

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(i) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation $x^{2}-4=0$
$x^{2}-4=0\\ x^{2}=4\\ x=\pm \sqrt{4}\\ x=2 \: \: \: \: \: \: \: x=-2$

Here – 2, 2 are the two roots of the equation.

Hence it is false that every quadratic equation has exactly one root.

(ii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation $x^{2}-x+2=0$
$x^{2}-x+2=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

$a=1,b=-1,c=2\\ b^{2}-4ac=(-1)^{2}-4(1)(2)\\ =1-8\\ =-7\\ b^{2}-4ac<0$

Hence both the roots of the equation are imaginary.

Hence the statement every quadratic equation has at least one real root is False.

(iii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation $x^{2}-4x+4=0$

$x^{2}-4x+4=0\\ x^{2}+(2)^{2}-2(x)(2)\\ (x-2)^{2}=0\: \: \: \: \: \: \: \: \: \: \: \: \: \: (using\: (a-b)^{2}=a^{2}+b^{2}-2ab)\\ (x-2)=0\\ x=2$

The equation $x^{2}-4x+4=0$ has only one root which is x = 2

Hence the given statement is False

(iv) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

As we know that the standard form of a quadratic equation is $ax^{2} + bx + c = 0$

It is a polynomial of degree 2.

As per the power of x is 2. There is at most 2 values of x exist that satisfy equation.

Hence the given statement every quadratic equations has at most two roots is true.

(v) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + x - 2 = 0$

$x^{2} + x - 2 = 0$ is a quadratic equation in which coefficient of x² and constant term have opposite signs.

$x^{2} + x - 2 = 0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$a=1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(1)(-2)\\ =1+8\\ =9\\ b^{2}-4ac>0$

Here $b^{2}-4ac>0$

Hence these types of equations have real roots.

(vi) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$ …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + 4= 0$

In $x^{2} + 4= 0$ , the coefficient of x² and constant term has same sign and coefficient of x is 0.

$x^{2} + 4= 0$

Compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$a=1,b=0,c=4\\ b^{2}-4ac=(0)^{2}-4(1)(4)\\ =16\\ b^{2}-4ac<0$

Here $b^{2}-4ac<0$

Hence these types of equations have no real roots.

Posted by

infoexpert24

View full answer

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Answers (2)

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