To find: A quadratic equation whose two solutions are 4 and 5

Question

Wriedge · Accepted Answer

Step 1
An equation of degree 2 is called quadratic equation. A quadratic equation always has two solutions (as number of solutions equal to degree of the equation).
Let

x^{2} + b x + c = 0

be a quadratic equation. When expression

x^{2} + b x + c

is factorized, always two factors are obtained. Suppose the factors are

(x - p)

and

(x - q)

Then by definition of the factors the expression is equal to the product of its factors, i.e.,
1)

x^{2} + b x + c = (x - p) \times (x - q) = x^{2} - (p + q) x + p \times q

From equation (1) it is clear:
(i) p and q are the solution of equation

x^{2} + b x + c = 0

, as solutions of an equation are obtained by putting factors equals zero, and
(ii)

b = - (p + q)

and

c = p \times q

The solutions of the required equation are 4 and 5. Therefore, we have

b = - (4 + 5) = - 9

and

c = p \times q = 4 \times 5 = 20

The required equation is

x^{2} + b x + c = x^{2} + (- 9) x + 20 = x^{2} - 9 x + 20 = 0

We also can use two formulae for finding equation.
1)

(x-first solution) \times (x-second solution) = 0 \Rightarrow (x - 4) (x - 5) = x^{2} - 9 x + 20 = 0

2)

x^{2} - (sum of roots) x + product of roots = x^{2} - (9) x + (20) = x^{2} - 9 x + 20 = 0

The quadratic equation whose solutions are 4 and 5 is

x^{2} - 9 x + 20 = 0

To find: A quadratic equation whose two solutions are 4 and 5

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