CBSE Class 10 Mathematics Quadratic Equations Notes Set 08

Standard Form and Zeros of a Quadratic Equation

A quadratic equation in the variable \( x \) has the standard form \( ax^2+bx+c=0 \), where \( a,b,c \) are real numbers and \( a\ne0 \). An equation such as \( (x+2)^2=2(x-3) \) is quadratic once it is expanded and simplified, because the highest power that remains is 2.
A value of \( x \) that makes the left side equal to zero is a root (or zero) of the equation. A quadratic equation has at most two roots. To test a number, substitute it and see whether the equation is satisfied.

Solving by Factorisation

Split the middle term so that the two parts have a product equal to \( ac \) and a sum equal to \( b \), factorise into two linear factors, and equate each factor to zero.
For \( 6x^2-x-2=0 \): \( ac=-12 \), and \( 3 \) and \( -4 \) have product \( -12 \) and sum \( -1 \). So \( 6x^2+3x-4x-2=3x(2x+1)-2(2x+1)=(3x-2)(2x+1)=0 \), giving \( x=\frac23 \) or \( x=-\frac12 \).

Quadratic Formula and Completing the Square

Completing the square on \( ax^2+bx+c=0 \) gives the quadratic formula \( x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \). It works for every quadratic equation, including those that do not factorise with integers.
Example: \( x^2-4x+1=0 \) gives \( x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt3 \).

Discriminant and Nature of Roots

The expression \( D=b^2-4ac \) is called the discriminant.
If \( D>0 \), there are two distinct real roots. If \( D=0 \), there are two equal real roots, each equal to \( -\frac{b}{2a} \). If \( D<0 \), there are no real roots.
Use these to find an unknown constant. For example, \( kx(x-2)+6=0 \) i.e. \( kx^2-2kx+6=0 \) has equal roots when \( 4k^2-24k=0 \), so \( k=6 \) (the value \( k=0 \) is rejected because the equation would no longer be quadratic).

Sum and Product of Roots

If \( \alpha \) and \( \beta \) are the roots of \( ax^2+bx+c=0 \), then \( \alpha+\beta=-\frac ba \) and \( \alpha\beta=\frac ca \).
A quadratic equation with roots \( \alpha,\beta \) is \( x^2-(\alpha+\beta)x+\alpha\beta=0 \).
Useful results: \( \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \) and \( \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \).

Equations Reducible to Quadratic Form

Equations with the variable in a denominator are cleared by multiplying through by the common denominator. Always check that no solution makes a denominator zero.
An equation like \( x+\frac5x=6 \) becomes \( x^2-6x+5=0 \), with solutions \( 1 \) and \( 5 \). When an expression repeats, substitute a new variable, for example \( t=\frac{x}{x+1} \) to get \( t+\frac1t=k \).

Word Problems

Choose a variable for the unknown quantity, translate the statement into an equation, solve it, and reject answers that are not meaningful (a negative length, age or speed).
For speed-time problems use \( \text{time}=\frac{\text{distance}}{\text{speed}} \); for upstream and downstream problems the speeds are \( v-s \) and \( v+s \), where \( v \) is the speed in still water and \( s \) is the speed of the stream. For work problems, add the rates \( \frac1x+\frac1y=\frac1t \).

Worked Example 1. Solve \( x+\frac{5}{x}=6 \).
Answer:
Multiply by \( x \): \( x^2-6x+5=0 \).
\( (x-1)(x-5)=0 \), so \( x=1 \) or \( x=5 \).

Worked Example 2. Find the roots of \( 2x^2-7x+3=0 \) by the formula.
Answer:
\( D=49-24=25 \).
\( x=\frac{7\pm5}{4} \), so \( x=3 \) or \( x=\frac12 \).

Worked Example 3. Find \( k \) if \( x^2+kx+9=0 \) has equal roots.
Answer:
\( D=k^2-36=0 \), so \( k=\pm6 \).

Worked Example 4. The sum of the squares of two consecutive positive integers is 365. Find them.
Answer:
Let them be \( x \) and \( x+1 \): \( x^2+(x+1)^2=365 \), so \( x^2+x-182=0 \).
\( (x+14)(x-13)=0 \), so \( x=13 \). The integers are 13 and 14.

Worked Example 5. A train covers 360 km at a uniform speed. At 5 km/h more it would take 1 hour less. Find the speed.
Answer:
\( \frac{360}{x}-\frac{360}{x+5}=1 \) gives \( x^2+5x-1800=0 \).
\( (x+45)(x-40)=0 \), so the speed is 40 km/h.