Quadratic Equations - MathsTips.com

An equation of the form $ax^2 + bx +c = 0 (a\neq 0)$ in which 2 is the highest power of x and a,b,c are any three numbers free from x is called an equation of second degree or a quadratic equation in x. Here a, b, c are constant terms; a is called the quadratic coefficient, b is called the linear coefficient and c is called the constant or free term. If a=0, then the equation is linear instead of quadratic.

Solving Quadratic Equation

A quadratic equation may be solved either by factorizing the left side( when the right side is zero) or by completing a square on the left side.

Example 1: Solve $6x^2+11x-10=0$

Solution: $6x^2 + 11x-10=0$

$\Rightarrow 6x^2 +15x-4x-10=0$

$\Rightarrow 3x(2x+5)-2(2x+5)=0$

$\Rightarrow (2x+5)(3x-5)=0$

Therefore, either $2x+5=0$ or, $3x-5=0$

Hence, $x= \dfrac{-5}{2} or x=\dfrac {2}{3}$

Example 2: Solve $2x^2-5x-2=0$

Solution: $2x^2-5x-2=0$

$\Rightarrow 4x^2 -10x=4$

$\Rightarrow (2x)^2-2(2x)(\dfrac{5}{2}) +(\dfrac {5}{2})^2 -\dfrac{25}{4}=4$

$\Rightarrow (2x-\dfrac{5}{2})^2=4+\dfrac{25}{4}= \dfrac{41}{4}$

$\Rightarrow 2x-\dfrac{5}{2}=\pm\sqrt{\dfrac{41}{4}} =\pm\dfrac{\sqrt{41}}{2}$

$\Rightarrow 2x=\dfrac{5}{2}\pm\dfrac{\sqrt{41}}{2}=\dfrac{5\pm\sqrt{41}}{2}$

$\Rightarrow x=\dfrac{5\pm\sqrt{41}}{4}$

Formula to Solve a Quadratic Equation:

The roots of the quadratic equation $ax^2 + bx +c = 0 (a\neq 0)$ is given by the following formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

This formula is known as Sridhar Acharya’s formula.

Example3: Solve $2x^2-10x+13=0$

Solution: According to Sridhar Acharya’s Formula, here, a=2, b=(-10), c=13

$\therefore$ $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4.2.13}}{2.2}$

$=\dfrac{10\pm\sqrt{100-104}}{4}$

$\therefore$ $x=\dfrac{10\pm\sqrt{-4}}{4} =\dfrac{10\pm2i}{4}=\dfrac{5\pm i}{2}$ which are complex roots. To know more about complex numbers and ‘i’ refer Complex Numbers.

Equations Reducible to Quadratic Form

Many equations does not look like quadratic equations but can be reduced to quadratic form very easily. Let us see some examples.

Example 4: Solve $2^{2x+3}+2^{x+3}=1+2^x$

Solution: $2^{2x+3}+2^{x+3}=1+2^x$

or, $2^{2x}.2^3+2^x.2^3=1+2^x$

or, $8.2^{2x}+8.2^x=1+2^x$ …(1)

Now, let $2^x=y$ Then, $y^2=(2^x)^2= 2^{2x}$

$\therefore$ from (1) we get,

$8y^2+8y=1+y$

or, $8y^2+7y-1=0$

or, $8y^2+8y-y-1=0$

or, $8y(y+1) -1(y+1)=0$

or, $(y+1)(8y-1)=0$

$\therefore$ $y=-1$

or $\dfrac{1}{8}$

If, $y=-1$ , then, $2^x=-1$ , which is not possible.

If, $y=\dfrac{1}{8}$ , then $2^x=\dfrac{1}{8}$

or, $2^x=\dfrac{1}{2^3}=2^{-3}$

$\therefore$ $x=-3$

Example 5: Solve $\sqrt{\dfrac{x}{1-x}} +\sqrt{\dfrac{1-x}{x}}=\dfrac{13}{6}$

Solution: Let us put $y=\sqrt{\dfrac{x}{1-x}}$ , then $\dfrac{1}{y}=\sqrt{\dfrac{1-x}{x}}$

$\therefore$ $y+\dfrac{1}{y} =\dfrac{13}{6}$ or, $\dfrac{y^2+1}{y}=\dfrac{13}{6}$

$\Rightarrow 6y^2+6=13y \\ \Rightarrow 6y^2-13y+6=0 \\ \Rightarrow 6y^2-4y-9y+6=0 \\ \Rightarrow (3y-2)(2y-3)=0$

$\therefore$ $y=\dfrac{2}{3}$ or, $y=\dfrac{3}{2}$

If $y=\dfrac{2}{3}, then \sqrt{\dfrac{x}{1-x}}=\dfrac{2}{3}$

$\Rightarrow \dfrac{x}{1-x} =\dfrac{4}{9} \\ \Rightarrow 9x=4-4x\\ \Rightarrow 13x=4$

$\therefore$ $x=\dfrac{4}{13}$

If $y=\dfrac{3}{2}, then \sqrt{\dfrac{x}{1-x}}=\dfrac{3}{2}$

$\Rightarrow \dfrac{x}{1-x} =\dfrac{9}{4} \\ \Rightarrow 4x=9-9x\\ \Rightarrow 13x=9$

$\therefore$ $x=\dfrac{9}{13}$

Answer: $x=\dfrac{4}{13} or, \dfrac{9}{13}$

Problems Leading to Quadratic Equations

Example 7: The sum of the squares of two numbers is 233 ad one of the numbers is 3 less than twice the other. Find the numbers.

Solution: Let one of the numbers be taken as x.

$\therefore$ the other number=(2x-3)

By the problem,

$x^2+(2x-3)^2=233 \\ \Rightarrow x^2+4x^2-12x+9=233 \\ \Rightarrow 5x^2-12x-224=0 \\ \Rightarrow 5x^2-40x-28x-224=0\\ \Rightarrow 5x(x-8)+28(x-8)=0\\ \Rightarrow (x-8)(5x+28)=0$

$\therefore$ $x=8$ or, $x=(-\dfrac{28}{5})$

If $x=8$ , then the other number is $(2 \times 8-3)=13$

If $x=\dfrac{-28}{5}$ then, the other number is $(2 \times \dfrac{-28}{5}-3)=\dfrac{-71}{5}$

Answer: The required numbers are either $8 and 13$ or, $\dfrac{-28}{5} and \dfrac{-71}{5}$ .

Sum and Product of Roots of a Quadratic Equation

If $\alpha \: \text{and} \: \beta$ be the roots of the quadratic equation $ax^2 + bx +c = 0 (a\neq 0)$ then, $\alpha +\beta=\dfrac{-b}{a}$ and $\alpha\beta=\dfrac{c}{a}$

From these two relations we obtain the following results:

If the two roots $\alpha \: \text{and} \: \beta$ be reciprocal to each other, then, $a=c$
If the two roots be equal in magnitude and opposite in sign then $b=0$

Example 8: If the roots of the equation $x^2-px+q=0$ be in the ration 2:3, prove that $6p^2=25q$ .

Solution: Let the roots of $x^2-px+q=0$ be $2 \alpha$ and $3 \alpha$ .

$\therefore$ $2 \alpha+3 \alpha=-\dfrac{(-p)}{1}=p$ …(1)

and, $2 \alpha.3 \alpha=q$ …(2)

From (1), $5 \alpha =p$ or, $\alpha=\dfrac{p}{5}$

From (2), $6\alpha^2=q \Rightarrow 6(\dfrac{p}{5})^2=q$

or, $6.\dfrac{p^2}{25}= q \Rightarrow 6p^2=25q$ (Proved)

Example 9: If the roots of the equation $x^2+px+7=0$ are denoted by $\alpha$ and $\beta$ and $\alpha^2 +\beta^2=22$ , find the value of p.

Solution: $\alpha +\beta=\dfrac{-p}{1} =(-p)$ and $\alpha \beta=\dfrac{7}{1}=7$

Now, $\alpha^2 + p^2= 22$ (given)

or, $(\alpha+\beta)^2-2 \alpha \beta=22$

or, $(-p)^2-2(7)=22$

or, $p^2=22+14=36$

$\therefore$ $p=\pm6$

Nature of Roots of a Quadratic Equation

The nature of the roots of a quadratic equation is determined by $(b^2-4ac)=D$ which is known as the discriminant of the quadratic equation.

Case 1: If D is positive, then the roots are real and unequal.
Case 2: If D is a perfect sqaure and a,b,c are all rational numbers, then the two roots are real, rational and unequal.
Case 3: If D is positive, but not a perfect square, then $\sqrt{D}$ is real and irrational. In this case the roots are real, irrational and unequal.
Case 4: If D=0, then the two roots are real and equal.
Case 5: If D is negative, then the roots are imaginary or complex. [Refer to Example 3 of this chapter]

Example 10: Prove that the equation $(x-a)(x-b) +(x-b)(x-c)+(x-a)(x-c)=0$ will have equal roots if and only if, $a=b=c$ .

Solution: $(x-a)(x-b) +(x-b)(x-c)+(x-a)(x-c)=0 \\ \Rightarrow 3x^2-2(a+b+c)+(ab+bc+ca)=0$

which is of the form, $Ax^2+Bx+C=0$ where $A=3, B={-2(a+b+c)}, C=ab+bc+ca$

For the given equation to have equal roots we must have,

$B^2-4AC=0$

$\Rightarrow {(-2)^2(a+b+c)}^2-4.3.(ab+bc+ca)=0$

$\Rightarrow 4(a^2+b^2+c^2+2ab+2bc+2ca)-12(ab+bc+ca)=0$

$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$

$\Rightarrow 2a^2+2B^2+2c^2-2ab-2bc-2ca=0$

$\Rightarrow (a^2+b^2-2ab)+(c^2+a^2-2ca)+(b^2+c^2-2bc)=0$

$\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$

$\therefore$ $(a-b)^2=0 \Rightarrow a=b$

$\therefore$ $(b-c)^2=0 \Rightarrow b=c$

$\therefore$ $(c-a)^2=0 \Rightarrow c=a$

Hence, $a=b=c$

Formation of a Quadratic Equation with Given Roots

Any quadratic equation can be written as,

$x^2-(Sum of the roots)x+(Product of the roots)=0$

Example 11: If $\alpha$ and $\beta$ be tyhe roots of the equation $x^2-px+q=0$ , form the equation where roots are $(\alpha \beta+\alpha+\beta)$ and $(\alpha \beta-\alpha-\beta)$

Solution: $\alpha +\beta= -\dfrac{(-p)}{1}$ and $\alpha \beta=\dfrac{q}{1}$

Sum of the roots of the required equation $=(\alpha \beta+\alpha+\beta) +(\alpha \beta-\alpha-\beta)=2 \alpha \beta=2q$

Product of the roots of the required equation $=(\alpha \beta+ \alpha+ \beta)(\alpha \beta- \alpha-\beta)= (\alpha \beta)^2-(\alpha+\beta)^2=q^2-p^2$

Hence the required equation is $x^2-2qx+(q^2-p^2)$

Conjugate Roots

Surd roots and complex roots of a quadratic equation always occur in conjugate pairs.

Example 12: Find the quadratic equation with real coefficients with one root: i) $(1+\sqrt{3})$ ii) $(3-2i)$

Solution: i) Since the quadratic equation with real coefficients has a root $(1+\sqrt{3})$ and surd roots always occur in pairs, the other root is $(1-\sqrt{3})$

Sum of the roots $= (1+\sqrt{3})+(1-\sqrt{3})=2$

Product of the roots $=(1+\sqrt{3})(1-\sqrt{3})=1^2-(\sqrt{3})^2=1-3=-2$

Hence the required equation is: $x^2-2x+(-2)=0$ or, $x^2-2x-2=0$

ii)Since one root is $(3-2i)$ and complex roots always occur in pairs, the other root is $(3+2i)$

Sum of the roots $= (3+2i)+(3-2i)=6$

Product of the roots $=(3-2i)(3+2i_= 3^2-(2i)^2=9+4=13$

Hence the required equation is: $x^2-6x+13=0$

Common Roots

Example 13: Find those values of k for which the equations $x^2-kx-21=0$ and $x^2-34x+35=0$ have a common root.

Solution: Let $\alpha$ be the common root of the given equations.

Then, $\alpha^2-k \alpha-21=0$ …(1)

and $\alpha^2-34 \alpha +35=0$ …(2)

Subtracting (2) from (1) we get, $2k \alpha-56=0$ or, $k \alpha=28$ or, $\alpha=\dfrac{28}{k}$

Substituting $\alpha=\dfrac{28}{k}$ in (1) we get,

$(\dfrac{28}{k})^2-k.\dfrac{28}{k}-21=0$

or, $\dfrac{(28)^2}{k^2}=49$

or, $k^2=\dfrac{28 \times 28}{49}=16$

or, $k=\pm4$

Exercise

The sum of the squares of two positive numbers is 232 and one of them is 4 less than thrice the other. Find the numbers.
Solve by completing the square: $3x^2+12x+8=0$
Comment on the nature of the roots of the equation $-17\dfrac{1}{2}x^2+42x=-3\dfrac{1}{2}$
Form the quadratic equation which has the roots: a) $(-1),(-2\dfrac{2}{3})$ b) $\dfrac{2\pm6i}{5}$
Solve: $x(x+1)(x+2)(x+3)=24$
If be the roots of the equation , find the value of:
1. $\alpha^2+ \beta^2$
2. $\dfrac{\alpha^2}{\beta} +\dfrac{\beta^2}{\alpha}$
If be the roots of the equation form an equation whose roots are:
1. $(\alpha+1)$ and $(\beta+1)$
2. $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$
Find the value of for which the equation will have:
1. Equal roots
2. Reciprocal roots
3. Roots whose product is 9
If the roots of the equation $ax^2+bx+c=0$ be in the ratio $4:5$ show that, $20b^2=81ac$
Find the equation with real coefficients whose one root is $(2-3i)$
If the equations $3x^2+px+1=0$ and $2x^2+px+1=0$ have a common root, show that $2p^2+3p^2-5pq+1=0$