Quadratic Equations Concepts

1. Quadratic Equation & Root Nature

• Formula:

$$ax^2+bx+c=0,\qquad \Delta=b^2-4ac$$ $$\Delta>0\Rightarrow\text{real and distinct roots}$$ $$\Delta=0\Rightarrow\text{real and equal roots}$$ $$\Delta<0\Rightarrow\text{non-real/complex roots}$$

Use the discriminant whenever the question asks about the nature or number of roots.

• Example: For what values of (k) does (x^2+kx+9=0) have equal roots? Solution:

For equal roots:

$$\Delta=0$$

Here:

$$a=1,\quad b=k,\quad c=9$$

Therefore:

$$k^2-4(1)(9)=0$$ $$k^2-36=0$$ $$(k-6)(k+6)=0$$ $$\boxed{k=\pm6}$$

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2. Quadratic Equation with Real/Distinct Roots

• Formula:

$$\boxed{\Delta>0}$$

For two distinct real roots, the discriminant must be positive.

• Example: Find the range of (k) for which (x^2-4x+k=0) has two distinct real roots. Solution:

$$\Delta=(-4)^2-4(1)(k)$$ $$=16-4k$$

For distinct real roots:

$$16-4k>0$$ $$k<4$$ $$\boxed{k<4}$$

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3. Quadratic Formula

• Formula:

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

Use when the actual roots are required and factorization is inconvenient.

• Example: Solve (2x^2-5x-3=0). Solution:

Here:

$$a=2,\quad b=-5,\quad c=-3$$ $$x=\frac{5\pm\sqrt{25+24}}{4}$$ $$=\frac{5\pm7}{4}$$

Thus:

$$x=\frac{12}{4}=3$$

or

$$x=\frac{-2}{4}=-\frac12$$ $$\boxed{x=3,-\frac12}$$

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4. Sum and Product of Roots

• Formula: If (\alpha,\beta) are roots of

$$ax^2+bx+c=0,$$

then:

$$\boxed{\alpha+\beta=-\frac ba}$$ $$\boxed{\alpha\beta=\frac ca}$$

Use these instead of solving the quadratic when only expressions involving the roots are required.

• Example: If (2x^2-5x+3=0) has roots (\alpha,\beta), find (\alpha+\beta) and (\alpha\beta). Solution:

$$\alpha+\beta=-\frac{-5}{2} =\boxed{\frac52}$$ $$\alpha\beta=\frac32$$ $$\boxed{\alpha\beta=\frac32}$$

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5. Expressions Involving Roots

• Formula:

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$ $$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$$ $$\frac1\alpha+\frac1\beta =\frac{\alpha+\beta}{\alpha\beta}$$

• Example: The roots of (2x^2-5x+3=0) are (\alpha,\beta). Find (\alpha^2+\beta^2). Solution:

$$\alpha+\beta=\frac52,\qquad \alpha\beta=\frac32$$

Therefore:

$$\alpha^2+\beta^2 ================ \left(\frac52\right)^2 -2\left(\frac32\right)$$ $$=\frac{25}{4}-3 =\frac{13}{4}$$ $$\boxed{\frac{13}{4}}$$

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6. Forming a Quadratic from Given Roots

• Formula: If roots are (\alpha,\beta):

$$\boxed{x^2-(\alpha+\beta)x+\alpha\beta=0}$$

Use when roots or their sum and product are known.

• Example: Form the quadratic equation whose roots are 3 and 5. Solution:

$$\alpha+\beta=8$$ $$\alpha\beta=15$$

Therefore:

$$x^2-8x+15=0$$ $$\boxed{x^2-8x+15=0}$$

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7. Transformed Roots

• Formula: If (\alpha,\beta) are roots and new roots are expressions such as (k\alpha,k\beta), use their sum and product:

$$k\alpha+k\beta=k(\alpha+\beta)$$ $$(k\alpha)(k\beta)=k^2\alpha\beta$$

For roots (\alpha+k,\beta+k):

$$\text{sum}=\alpha+\beta+2k$$ $$\text{product}=\alpha\beta+k(\alpha+\beta)+k^2$$

• Example: If (\alpha,\beta) are roots of (x^2-5x+6=0), form the equation whose roots are (2\alpha,2\beta). Solution:

Original:

$$\alpha+\beta=5,\qquad\alpha\beta=6$$

New sum:

$$2\alpha+2\beta=10$$

New product:

$$(2\alpha)(2\beta)=24$$

Therefore:

$$\boxed{x^2-10x+24=0}$$

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8. Common Root of Two Quadratics

• Formula: A common root (\alpha) must satisfy both equations:

$$a_1\alpha^2+b_1\alpha+c_1=0$$ $$a_2\alpha^2+b_2\alpha+c_2=0$$

Subtract/eliminate terms to obtain the common root or parameter.

• Example: If (x^2-px+12=0) and (x^2-7x+12=0) share a root, find (p). Solution:

The second equation factors as:

$$x^2-7x+12=(x-3)(x-4)$$

So its roots are (3,4).

If the first equation shares a root, test:

For (x=3):

$$9-3p+12=0$$ $$3p=21$$ $$p=7$$

For (x=4):

$$16-4p+12=0$$ $$4p=28$$ $$p=7$$

Therefore:

$$\boxed{p=7}$$

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9. Common Both Roots

• Formula: Two quadratic equations have the same two roots if their coefficients are proportional:

$$\boxed{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}}$$

For monic quadratics, equal roots sets imply equal coefficients.

• Example: Find (k) if (x^2-6x+8=0) and (2x^2-kx+16=0) have the same roots. Solution:

For the equations to represent the same roots, the second must be twice the first:

$$2(x^2-6x+8) =2x^2-12x+16$$

Comparing:

$$k=12$$ $$\boxed{k=12}$$

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10. Completing the Square

• Formula:

$$ax^2+bx+c ========= a\left(x+\frac{b}{2a}\right)^2 +\frac{4ac-b^2}{4a}$$

For (a>0), the expression has a minimum; for (a<0), it has a maximum.

• Example: Find the minimum value of (x^2-6x+13). Solution:

$$x^2-6x+13$$ $$=(x-3)^2+4$$

Since:

$$(x-3)^2\geq0$$

Minimum occurs when (x=3).

$$\boxed{\text{Minimum value}=4}$$

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11. Maximum/Minimum Value Using Vertex

• Formula: For

$$f(x)=ax^2+bx+c,$$

the turning point occurs at:

$$\boxed{x=-\frac{b}{2a}}$$

and:

$$\boxed{f_{\text{extreme}}=\frac{4ac-b^2}{4a}}$$

Use minimum when (a>0), maximum when (a<0).

• Example: Find the maximum value of (-2x^2+8x+5). Solution:

$$a=-2,\quad b=8,\quad c=5$$ $$x=-\frac{8}{2(-2)}=2$$

Substitute:

$$f(2)=-2(4)+16+5$$ $$=-8+21=13$$ $$\boxed{\text{Maximum}=13}$$

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12. Number of Positive/Negative Roots

• Formula: For roots (\alpha,\beta):

$$\alpha\beta=\frac ca$$ $$\alpha+\beta=-\frac ba$$

If (\alpha\beta<0), roots have opposite signs. If (\alpha\beta>0), roots have the same sign; use the sum to determine whether both are positive or negative.

• Example: Determine the signs of the roots of (x^2-7x+12=0). Solution:

$$\alpha+\beta=7>0$$ $$\alpha\beta=12>0$$

Product positive means same sign.

Sum positive means both are positive.

$$\boxed{\text{Both roots are positive}}$$

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13. Positive and Negative Root Conditions

• Formula: For

$$ax^2+bx+c=0,$$

two real roots of opposite signs require:

$$\boxed{\frac ca<0}$$

Two positive roots require:

$$\boxed{\Delta\geq0,\quad -\frac ba>0,\quad \frac ca>0}$$

Two negative roots require:

$$\boxed{\Delta\geq0,\quad -\frac ba<0,\quad \frac ca>0}$$

• Example: For what values of (k) does (x^2-kx+4=0) have two positive real roots? Solution:

For real roots:

$$k^2-16\geq0$$ $$|k|\geq4$$

For positive roots:

$$\alpha+\beta=k>0$$

Hence:

$$k\geq4$$

For distinct positive roots:

$$\boxed{k>4}$$

For positive real roots allowing equal roots:

$$\boxed{k\geq4}$$

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Advanced Variants

14. Parameter for Roots in a Given Interval

• Formula: For a quadratic (f(x)), to ensure both roots lie in an interval, combine:

$$\Delta\geq0$$

with sign/value conditions at the interval boundaries and the vertex.

• Example: Find the condition on (k) such that both roots of (x^2-kx+4=0) are positive. Solution:

For both roots to be positive:

$$\alpha+\beta=k>0$$

and:

$$\alpha\beta=4>0$$

Also:

$$\Delta=k^2-16\geq0$$

Thus:

$$k\geq4$$ $$\boxed{k\geq4}$$

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15. Quadratic Inequality

• Formula: First find the roots of:

$$ax^2+bx+c=0$$

then use the sign of (a) to determine the intervals.

For (a>0):

$$ax^2+bx+c>0$$

outside the roots and (<0) between them.

• Example: Solve (x^2-5x+6>0). Solution:

Factor:

$$x^2-5x+6=(x-2)(x-3)$$

Roots:

$$x=2,3$$

Since the coefficient of (x^2) is positive, the expression is positive outside the roots.

Therefore:

$$\boxed{x<2\text{ or }x>3}$$

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16. Minimum/Maximum of a Quadratic Under a Constraint

• Formula: If (x+y=k), substitute:

$$y=k-x$$

into the expression and reduce it to a single-variable quadratic. Then use the vertex/minimum formula.

• Example: If (x+y=10), find the minimum value of (x^2+y^2). Solution:

$$y=10-x$$

Therefore:

$$x^2+y^2=x^2+(10-x)^2$$ $$=2x^2-20x+100$$

Complete the square:

$$=2(x-5)^2+50$$

Minimum occurs at:

$$x=5,\quad y=5$$

Hence:

$$\boxed{\text{Minimum}=50}$$

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17. Reciprocal Roots

• Formula: If (\alpha,\beta) are roots of

$$ax^2+bx+c=0,$$

then (1/\alpha,1/\beta) have:

$$\text{sum}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-b}{c}$$ $$\text{product}=\frac1{\alpha\beta}=\frac ac$$

• Example: Find the quadratic whose roots are the reciprocals of the roots of (2x^2-5x+3=0). Solution:

Original:

$$\alpha+\beta=\frac52,\qquad\alpha\beta=\frac32$$

For reciprocal roots:

\frac1\alpha+\frac1\beta ======================== # \frac{5/2}{3/2} \frac53 $$\frac1{\alpha\beta}=\frac23$$

Thus:

$$x^2-\frac53x+\frac23=0$$

Multiplying by 3:

$$\boxed{3x^2-5x+2=0}$$