Question

If 2 and -5 are the roots of the equation then the equation is

Answer 1

Answer:

$x^{2}$ + 3$x$ - 10

Step-by-step explanation:

Just remember this : when two roots are given, take one as α and other as β

Now the equation given or rather a formula for finding the equation is :

$x^{2}$ - (α+β)$x$ + αβ

Now replace the α and β with the two roots.

Here I take α = 2 and β = -5

so the equation forms as follows:

$x^{2}$ - (α+β)$x$ + αβ

⇒ $x^{2}$ - (2+(-5))$x$ + 2(-5)

⇒ $x^{2}$ - (-3)$x$ - 10

⇒ $x^{2}$ + 3$x$ +10

Hope it helps

Answer 2

Answer :

x² + 3x - 10 = 0

Note :

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; ax² + bx + c = 0

★ If α and ß are the roots of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots , (α + ß) = -b/a

• Product of roots , (αß) = c/a

★ If α and ß are the roots of a quadratic equation , then that quadratic equation is given as : k•[ x² - (α + ß)x + αß ] = 0 , k ≠ 0.

Solution :

Here ,

The given roots of the required quadratic equation are : 2 and -5 .

Let α = 2 and ß = -5 .

Now ,

• Sum of the roots will be given as ;

α + ß = 2 + (-5) = 2 - 5 = -3

• Product of the roots will be given as ;

αß = 2•(-5) = -10

Now ,

The required quadratic equation will be given as ;

=> k•[ x² - (α + ß)x + αß ] = 0 , k ≠ 0

=> k•[x² - (-3)x + (-10)] = 0 , k ≠ 0

=> k•[x² + 3x - 10] = 0 , k ≠ 0

If k = 1 , then the given quadratic equation will be ; x² + 3x - 10 = 0 .