Quadratic Equation
Category : 10th Class
QUADRATIC EQUATION
FUNDAMENTALS
Note: (i) An equation of degree 2 is called a quadratic equation.
(ii) The quadratic equation is of the form\[a{{x}^{2}}+bx+c=0\].
Note: (i) Since the degree of a quadratic equation is 2, if has 2 roots or solutions.
(ii) \[x=a\]is the root of \[p\left( x \right)=0,\]if \[p\left( a \right)=0.\]
(iii) Finding the roots of a quadratic equation is called solving the quadratic equation.
(a) Factorization method by:
(i) Splitting the middle term (ii) Completing the square
(b) Using formula for roots \[(\alpha ,\beta )=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Step 1: Find the product of the coefficient of \[{{x}^{2}}\] and the constant term i.e., ac.
Step 2: (a) If ac is positive, then choose two factors of ac, whose sum is equal to b (the coefficient of the middle term).
(b) If ac is negative, then choose two factors of ac, whose difference is equal to b (the coefficient of the middle term).
Step 3: Express the middle terms as the sum (or difference) of the two factors obtained in step 2. [Now the quadratic equation has 4 terms]
Step 4: Express the given quadratic equation as a product of two binomials, and solve them. The two values obtained in step 2 are the roots of the given quadratic equation.
Elementary question - 1: Solve \[{{x}^{2}}+7x+12=0\]
Answer; ac = 1.12 = 12: Factors are 1, 2, 3, 4, 6, 12
The combination which gives middle term b =7 is 3 + 4 = 7
\[\therefore \] Written as \[{{x}^{2}}+3x+4x+12=0\]
or \[x\left( x+3 \right)+4\left( x+3 \right)=0\text{ }or\text{ }\left( x+4 \right)\left( x+3 \right)=0\]
(ii) Completing the square: In some cases, where the given quadratic equation can be solved by factorization, a suitable term is added and subtracted. Then terms are regrouped in such a manner that a square is completed by three of the terms. The equation is then solved using factorization method.
Usually, the term added and subtracted is the square of half the coefficient of x.
Note: The roots of the quadratic equation using the quadratic formula are \[x=\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[x=\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]
Roots of Biquadratic Equation
Any biquadratic equation, \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0,\] will have four roots. If a, p, y and 5 are its roots, the relation between the roots is given by
Sum of the roots \[=\alpha +\beta +\gamma +\delta =-\frac{b}{a}\]
Sum of product of two roots at a time \[=\alpha \beta +\beta \gamma +\gamma \delta +\delta \alpha =\frac{c}{a}\]
Sum of product of three roots at a time \[=\alpha \beta \gamma +\beta \gamma \delta +\gamma \delta \alpha +\delta \alpha \beta =-\frac{d}{a}\]
Product of the roots \[=\alpha \beta \gamma \delta =\frac{e}{a}\]
Constructing a New Quadratic Equation by Changing the Roots of a Given Quadratic Equation
If we are given a quadratic equation, we can build a new quadratic equation by changing the roots of this equation in the manner specified to us.
For example, consider the quadratic equation \[a{{x}^{2}}+bx+c=0\] and let its roots be \[\alpha \] and \[\beta \] respectively. Then, we can build new quadratic equations as per the following points.
Finding the roots of a quadratic equation by Graphical Method
Method
We can also solve the quadratic equation \[p{{x}^{2}}+qx+r=0\]by considering the following equation: Clearly, \[y=p{{x}^{2}}\]is a parabola and \[y=-qx-r\]is a straight line
Step 1: Draw the graph of \[y=p{{x}^{2}}\]and \[y=-qx-r\]on the same graph paper.
Step 2: Draw perpendiculars from the points of intersection of parabola and the straight line onto the X-axis. Let the points of intersection on the X-axis be \[\left( {{x}_{1}},0 \right)\] and\[\left( {{x}_{2}},0 \right)\].
Step 3: The x-coordinates of the points in Step (2), i.e., \[{{x}_{1}}\] and \[{{x}_{2}}\] are the two distinct roots of \[p{{x}^{2}}+px+r=0\]
Example:
Solve \[2{{x}^{2}}-x-3=0\]
We know that the roots of \[2{{x}^{2}}-x-3=0\] are the x-coordinates of the points of intersection of the parabola, \[y=2{{x}^{2}}\]and the straight line, \[y=x+3\]
(1) |
\[y=2{{x}^{2}}\] |
||||
x |
0 |
1 |
2 |
- 1 |
-2 |
\[y=2{{x}^{2}}\] |
0 |
2 |
8 |
2 |
8 |
(2) |
\[y=x+3\] |
|||||
x |
0 |
1 |
2 |
-1 |
-2 |
-3 |
\[y=x+3\] |
3 |
4 |
5 |
2 |
1 |
0 |
Draw the graph of \[y=2{{x}^{2}}\] and \[y=x+3\](see Fig.)
Clearly, the perpendiculars drawn from the points of intersection of parabola and the line meet the X-axis at\[\left( \frac{3}{2},0 \right)\] and \[(-1,0)\].
\[\therefore \] The roots of the given quadratic equation \[2{{x}^{2}}-x-3=0\] are \[\frac{3}{2}\] and – 1.
Note:
3. If the line does not meet the parabola, i.e., when the line and the parabola have no points in common, then the quadratic equation has no real roots. In this case, the roots of the quadratic equation are imaginary (see Figure)
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