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

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