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Last modified on August 3rd, 2023

The roots of a quadratic equation are the values of the variable that satisfies the equation. They are also known as the ‘zeroes’ of the quadratic equation. For the equation ax^{2} + bx + c = 0 the two roots α and β are:

- ${\alpha =\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}}$
- β = ${\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

For example, the two roots of the quadratic equation x^{2} + 5x + 6 = 0 are {-3, -2} because they satisfy the equation when their values are substituted.

When x = -3, (-3)^{2} + 5 × -(3) + 6 = 0

When x = -2, (-2)^{2} + 5 × -(2) + 6 = 0

**Without solving the equation 3x ^{2} – 2x – 1 = 0, find whether x = 1 is a root or solution of this equation.**

Solution:

Substituting x = 1 in the given equation 3x^{2} – 2x – 1 = 0, we get

3(1)^{2} – 2 (1) – 1 = 0

=> 3 – 2 – 1 = 0

=> 3 – 3 = 0

Therefore, x = 1 is a solution of the given equation 3x^{2} – 2x – 1 = 0

We can solve the quadratic equation to find its roots in different ways.

As we know α and β are the two roots of the quadratic equation, whose values can be determined using the quadratic formula:

(α, β) = ${\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Let us verify if, indeed, the roots of the quadratic equation x^{2} + 5x + 6 = 0 are {-3, -2}

In the equation x^{2} + 5x + 6 = 0, a = 1, b = 5, c = 6

(α, β) = ${\dfrac{-5\pm \sqrt{\left( 5\right) ^{2}-4\times 1\times 6}}{-2\times 1}}$

= ${\dfrac{-5\pm \sqrt{25-24}}{2}}$

= ${\dfrac{-5+\sqrt{1}}{2}}$

= ${\dfrac{-5+1}{2}}$ and ${\dfrac{-5-1}{2}}$

= {-3, -2}

Thus, the two roots of the quadratic equation x^{2} + 5x + 6 = 0 are {-3, -2}

Here we will factor the middle term ‘5x’ of the same equation x^{2} + 5x + 6 = 0

x^{2} + 5x + 6 = 0

=> x^{2} + 3x +2x + 6 = 0

=> x(x + 3) + 2(x + 3) = 0

=> (x + 3)(x + 2)= 0

=> x + 3 = 0 or x + 2 = 0

=> x = {-3, -2}

Thus, the two roots of the quadratic equation x^{2} + 5x + 6 = 0 are {-3, -2}

**Step 1**: Isolating the x^{2} and x terms to one side of the equation

y = x^{2} + 5x + 6

=> y – 6 = x^{2} + 5x

**Step 2**: Add ${\left(\dfrac{b}{2}\right) ^{2}}$ to both sides of the equation

Here, b = ${\left( \dfrac{5}{2}\right) ^{2}}$ = 25/4

=> ${y-6+\dfrac{25}{4}=x^{2}+5x+\dfrac{25}{4}}$

**Step 3**: Factor the right side of the equation into a perfect square

=> ${y+\dfrac{1}{4}=\left( x+\dfrac{5}{2}\right) ^{2}}$

**Step 4**: Rewriting the equation in terms of y

=> ${y=\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}}$

When the quadratic function f(x) = y = 0

${\left( x+\dfrac{5}{2}\right) ^{2}-\dfrac{1}{4}=0}$

=> ${\left( x+\dfrac{5}{2}\right) ^{2}=\dfrac{1}{4}}$

=> ${x+\dfrac{5}{2}=\pm \sqrt{\dfrac{1}{4}}}$

=> ${x=\left\{ +\dfrac{1}{2}-\dfrac{5}{2},-\dfrac{1}{2}-\dfrac{5}{2}\right\}}$

=> x= {-3, -2}

Thus, the two roots of the quadratic equation x^{2} + 5x + 6 = 0 are {-3, -2}

Graphing the quadratic equation manually using a graphing calculator and then identifying the x-intercepts will give the roots of the quadratic equation.

The graph shows the two x-intercepts are (-2, 0) and (-3, 0). Thus the two roots of the quadratic equation are (-3, -2)

The nature of the roots of the quadratic equation depends on the value of the discriminant as follows:

- If b
^{2}– 4ac > 0, the quadratic equation has 2 real solutions - If b
^{2}– 4ac < 0, the quadratic equation has no real solutions but two different complex or imaginary roots - If b
^{2}– 4ac = 0, the quadratic equation has one real root

**Find the value(s) of g if the quadratic equation 3y ^{2} + gy + 2 = 0 has equal roots.**

Solution:

As we know, thequadratic equation ax^{2} + bx + c = 0 has equal roots if its discriminant b^{2} – 4ac = 0

Here, a = 3, b = g, and c = 2

Putting the value in the discriminant, we get

b^{2} – 4ac = 0

=> (g)^{2} – 4 × 3 × 2 = 0

=> g^{2}– 24 = 0

=> g^{2} = 24

=>g = ${\pm 2\sqrt{6}}$

Thus, when the given quadratic equation has equal roots, g = ${2\sqrt{6}}$ or g = ${-2\sqrt{6}}$

The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.

α + β = ${-\dfrac{b}{a}}$

**Derivation**

α + β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}+\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

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

= ${-\dfrac{2b}{2a}}$ = ${-\dfrac{b}{a}}$

The product of the roots of a quadratic equation is equal to the constant term divided by the leading coefficient.

α.β = ${\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}×\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}}$

= ${\dfrac{b^{2}+b\sqrt{b^{2}-14ac}-b\sqrt{b^{2}-4ac}-\left( b^{2}-4ac\right) }{4a^{2}}}$

= ${\dfrac{b^{2}-b^{2}+4ac}{4a^{2}}}$

= ${\dfrac{4ac}{4a^{2}}}$ = ${\dfrac{c}{a}}$

**Find the sum and the product of the roots of the quadratic equation a ^{2} – 5a – 14 = 0**

Solution:

Comparing the given quadratic equation with the standard form ax^{2} + bx + c = 0, here a = 1, b = -5, c = -14

As we know,

α + β = ${-\dfrac{b}{a}}$

= 1

Thus the sum of the roots of the quadratic equation a^{2} – 5a – 14 = 0 is 1

α.β = ${\dfrac{c}{a}}$

= -14

Thus the product of the roots of the quadratic equation a^{2} – 5a – 14 = 0 is -14

**Write the quadratic equation with the given roots {7, -2}**

Solution:

The quadratic equation can be obtained by multiplying the factors formed by the roots {7, -2}

The factors are (x – 3)(x + 2)

Thus, the quadratic equation is (x – 3)(x + 2) = 0

=> x^{2} + 2x – 3x – 6 = 0

=> x^{2} – x – 6 = 0

Last modified on August 3rd, 2023