CBSE 10th Mathematics | Introduction of Quadratic Equation

Introduction of Quadratic Equation

A polynomial of degree 2 (i.e.\displaystyle a{{x}^{2}}+bx+c) is called a quadratic polynomial where a ¹ 0 and a, b, c are real numbers.

Any equation of the form, p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. Therefore, \displaystyle a{{x}^{2}}+bx+c=0,\,\,a\ne 0 is called the standard form of a quadratic equation.

e.g. \displaystyle 2{{x}^{2}}-3x+7, \displaystyle 8{{x}^{2}}+x-\sqrt{19}

Zeros of Quadratic Equation

For a quadratic polynomial \displaystyle p(x)=a{{x}^{2}}+bx+c, those values of x for which \displaystyle a{{x}^{2}}+bx+c=0 is satisfied, are called zeros of quadratic polynomial \displaystyle p(x), i.e. if \displaystyle p(\alpha )=a{{\alpha }^{2}}+b\alpha +c=0, then a is called the zero of quadratic polynomial.

Roots of Quadratic Equation

If a, b are zeros of polynomial \displaystyle a{{x}^{2}}+bx+c, then a, b are called roots (or solutions) of corresponding equation \displaystyle a{{x}^{2}}+bx+c=0 which implies that \displaystyle p(\alpha )=p(\beta )=0.

i.e., \displaystyle a{{\alpha }^{2}}+b\alpha +c=0 and \displaystyle a{{\beta }^{2}}+b\beta +c=0.

Problems based on Quadratic Equation:

Example:

Check whether the following equations are quadratic or not.

(i) \displaystyle {{(x-2)}^{2}}+1=2x-3    (ii) \displaystyle x(x+1)+8=(x+2)\,\,(x-2)

(iii) \displaystyle 7x-2{{x}^{2}}            (iv) \displaystyle x+\frac{1}{x}=2

Solution:

(i)    \displaystyle {{(x-2)}^{2}}+1=2x-3 can be rewritten as

\displaystyle {{x}^{2}}-4x+4+1=2x-3

\displaystyle {{x}^{2}}-4x+5=2x-3

i.e.,            \displaystyle {{x}^{2}}-6x+8=0

It is of the form \displaystyle a{{x}^{2}}+bx+c=0.

Therefore, the given equation is a quadratic equation.

(ii)    As \displaystyle x(x+1)+8={{x}^{2}}+x+8 and \displaystyle (x+2)\,(x-2)={{x}^{2}}-4

Therefore, we have        \displaystyle {{x}^{2}}+x+8={{x}^{2}}-4

Þ                 \displaystyle x+12=0

It is not of the form \displaystyle a{{x}^{2}}+bx+c=0.

Therefore, the given equation is not a quadratic equation.

(iii)    \displaystyle 7x=2{{x}^{2}}\,\,\,\Rightarrow \,\,2{{x}^{2}}-7x=0

As \displaystyle 2{{x}^{2}}-7x is a quadratic polynomial

\ \displaystyle 2{{x}^{2}}-7x=0 is a quadratic equation.

(iv)    \displaystyle x+\frac{1}{x}=2

Þ    \displaystyle x+{{x}^{-1}}-2=0

Since \displaystyle x+{{x}^{-1}}-2 is not a quadratic polynomial.

\ \displaystyle x+\frac{1}{x}=2 is not a quadratic equation

Example:

In each of the following, determine the value of k for which the given value is a solution of the equation.

(i) \displaystyle k{{x}^{2}}-3x+2=0;\,\,x=2        (ii) \displaystyle 2{{x}^{2}}+kx+6=0;\,\,x=-2

Solution:

(i)     \displaystyle k{{x}^{2}}-3x+2=0            …(i)

Since \displaystyle x=2 is a solution of the equation

Now substituting x = 2     in equation (i), we get

\displaystyle k{{(2)}^{2}}-3(2)+2=0

Þ    \displaystyle 4k-4=0

Þ    \displaystyle 4k=4

Þ    \displaystyle k=4\div 4=1

(ii)    Since x = –2 is a solution of the equation

\displaystyle 2{{x}^{2}}+kx+6=0            …(i)

Now substituting x = – 2 in (i), we get

\displaystyle 2{{(-2)}^{2}}+k(-2)+6=0

Þ    \displaystyle 8-2k+6=0

\displaystyle -2k=-14

or    \displaystyle 2k=14

Þ    \displaystyle k=14\div 2=7

Example:

In each of the following, determine whether the given values of x are the solutions of the given equation or not:

(i) \displaystyle {{x}^{2}}+6x+5=0;\,\,x=-1,\,\,x=-5        (ii) \displaystyle 6{{x}^{2}}-x-2=0;\,\,x=-\frac{1}{2},\,\,x=\frac{2}{3}.

Solution:

(i)    Consider \displaystyle p(x)=\displaystyle {{x}^{2}}+6x+5        …(i)

Substituting x = –1 in (i), we get

\displaystyle p(-1)= \displaystyle {{(-1)}^{2}}+6\times (-1)+5

= 1 – 6 + 5

= \displaystyle 6-6=0.

Hence x = 1 is the solution of given equation.

Again substituting x = –5 in (i), we get

\displaystyle p(-5)= \displaystyle {{(-5)}^{2}}+6\times (-5)+5

= 25 – 30 + 5

= 30 – 30 = 0

Hence x = 5 is the solution of given equation.

\    x = –1 and x = –5 are the solutions of given equation.

(ii)    Consider p(x)    \displaystyle =6{{x}^{2}}-x-2        …(i)

Substituting \displaystyle x=-\frac{1}{2} in (i), we get

\displaystyle p\left( -\frac{1}{2} \right)= \displaystyle 6\times {{\left( -\frac{1}{2} \right)}^{2}}-\left( -\frac{1}{2} \right)-2

= \displaystyle 6\times \frac{1}{4}+\frac{1}{2}-2=\frac{3}{2}+\frac{1}{2}-2 = \displaystyle \frac{3+1}{2}-2=\frac{4}{2}-2

= 2 – 2 = 0

Þ    \displaystyle x=-\frac{1}{2} is the solution of given equation.

Again substituting \displaystyle x=\frac{2}{3} in (i), we get

\displaystyle p\left( \frac{2}{3} \right)= \displaystyle 6\times {{\left( \frac{2}{3} \right)}^{2}}-\frac{2}{3}-2 = \displaystyle 6\times \frac{4}{9}-\frac{2}{3}-2=\frac{8}{3}-\frac{2}{3}-2=\frac{6}{3}-2=0

Þ         \displaystyle x=\frac{2}{3} is the solution of given equation.

\    \displaystyle x=-\frac{1}{2} and\displaystyle x=\frac{2}{3} are the solutions of quadratic equation.

Example:

Determine whether the given values of x are solutions of the given equation or not:

\displaystyle {{x}^{2}}+2\sqrt{3}x-9=0;\,\,x=\sqrt{3},\,\,x=-3\sqrt{3}

Solution:

Consider \displaystyle p(x)=\displaystyle {{x}^{2}}+2\sqrt{3}x-9        …(i)

Substituting \displaystyle x=\sqrt{3} in (i)

\displaystyle p\left( \sqrt{3} \right)= \displaystyle {{(\sqrt{3})}^{2}}+2\sqrt{3}(\sqrt{3})-9=3+6-9=0

\    \displaystyle x=\sqrt{3} is a solution of the given quadratic equation.

Substituting \displaystyle x=-3\sqrt{3} in (i) we get

\displaystyle p(-3\sqrt{3}) = \displaystyle {{(-3\sqrt{3})}^{2}}+2\sqrt{3}(-3\sqrt{3})-9

= 27 – 18 – 9 = 0 = RHS

\     \displaystyle x=-3\sqrt{3} is a solution of the given quadratic equation.

Example:

If x = 2 and x = 3 are roots of \displaystyle 3{{x}^{2}}-2px+2q=0, find the values of p and q.

Solution:

Since x = 2 is a root of the given equation,

\ we have \displaystyle 3{{(2)}^{2}}-2p(2)+2q=0

Þ     \displaystyle 12-4p+2q=0        Þ    \displaystyle 4p-2q=12        …(i)

Since x = 3 is a root of the given equation,

\ we have    \displaystyle 3{{(3)}^{2}}-2p(3)+2q=0

Þ     \displaystyle 27-6p+2q=0        Þ    \displaystyle 6p-2q=27        …(ii)

Subtracting equation (i) from equation (ii), we get

\displaystyle 6p-2q-4p+2q=27-12

or         \displaystyle 2p=15        Þ    \displaystyle p=\frac{15}{2}

Substituting \displaystyle p=\frac{15}{2} in equation (i), we get

\displaystyle 4\left( \frac{15}{2} \right)-2q=12

Þ    \displaystyle 2q=30-12     Þ    \displaystyle 2q=18 Þ    q = 9

Hence \displaystyle p=\frac{15}{2} and q = 9.

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