Question

A quadratic equation whose one root is 3 and the sum of
the roots is zero is zero find the quadratic equation
O
X^2+9=0
x^2-9=0
9x^2+1=0
9x^2-1=0​

Answer 1

Aɴsᴡᴇʀ :-

• $\sf\large\orange{ {x}^{2} - 9 = 0 }$

Gɪᴠᴇɴ :-

• $\sf\large\green{ \alpha = 3 \: ,\alpha + \beta = 0}$

Tᴏ Fɪɴᴅ :-

• The quadratic equation.

Sᴏʟᴜᴛɪᴏɴ :-

• Let α be the one root of the equation and ${\beta}$ be the other root.

$\sf\large\purple{ ➯\: \alpha + \beta = 0}$

$\sf\large\purple{ ➯\: 3 + \beta = 0}$

$\sf\large\purple{ ➯\: \beta = - 3}$

• $\sf\large{The \: value \: of \: \alpha = 3 \: and \: \beta = - 3........(1)}$

• $\sf\large{Product \: of \: zeroes\: = \alpha \times \beta }$

$\sf\large\purple{ ➯ \: \alpha \times \beta = 3 \times - 3 }$

$\sf\large\purple{ ➯\: \alpha \beta = - 9.........(2) }$

• If $\sf{ \alpha }$ and $\sf{ \beta }$ are the zeroes of the quadratic equation then quadratic equation will be :

$\sf\large\blue{ → [{x}^{2} - ( \alpha + \beta )x + \alpha \beta] }$

$\sf\large\blue{ → {x}^{2} - ( 3 - 3)x + ( - 9) }$

[ From equation 1 and 2 ]

$\sf\large\pink{→ {x}^{2} - 0 \times x - 9 }$

$\sf\large\pink{ → {x}^{2} - 9 = 0 }$

• Hence the quadratic equation is x² - 9 = 0.

Answer 2

Answer:

Aɴsᴡᴇʀ :-

\sf\large\orange{ {x}^{2} - 9 = 0 }x

2

−9=0

Gɪᴠᴇɴ :-

\sf\large\green{ \alpha = 3 \: ,\alpha + \beta = 0}α=3,α+β=0

Tᴏ Fɪɴᴅ :-

The quadratic equation.

Sᴏʟᴜᴛɪᴏɴ :-

Let α be the one root of the equation and {\beta}β be the other root.

\sf\large\purple{ ➯\: \alpha + \beta = 0}➯α+β=0

\sf\large\purple{ ➯\: 3 + \beta = 0}➯3+β=0

\sf\large\purple{ ➯\: \beta = - 3}➯β=−3

\sf\large{The \: value \: of \: \alpha = 3 \: and \: \beta = - 3........(1)}Thevalueofα=3andβ=−3........(1)

\sf\large{Product \: of \: zeroes\: = \alpha \times \beta }Productofzeroes=α×β

\sf\large\purple{ ➯ \: \alpha \times \beta = 3 \times - 3 }➯α×β=3×−3

\sf\large\purple{ ➯\: \alpha \beta = - 9.........(2) }➯αβ=−9.........(2)

If \sf{ \alpha }α and \sf{ \beta }β are the zeroes of the quadratic equation then quadratic equation will be :

\sf\large\blue{ → [{x}^{2} - ( \alpha + \beta )x + \alpha \beta] }→[x

2

−(α+β)x+αβ]

\sf\large\blue{ → {x}^{2} - ( 3 - 3)x + ( - 9) }→x

2

−(3−3)x+(−9)

[ From equation 1 and 2 ]

\sf\large\pink{→ {x}^{2} - 0 \times x - 9 }→x

2

−0×x−9

\sf\large\pink{ → {x}^{2} - 9 = 0 }→x

2

−9=0

Hence the quadratic equation is x² - 9 = 0.