Question

11. Determine the quadratic equation given the surn of the
roots -3 and product of the roots 7.
A. x2 – 3x + 7 = 0
B. x2 + 3x + 7 = 0
C. x2 - 3x - 7 = 0
D. x2 + 3x = 7
​

Answer 1

Given :

• Sum of the roots = - 3
• Product of roots = 7

To find :

• Determine the quadratic equation

Solution :

Here we have,

Sum of roots (α + β) = -3

Product of roots (αβ) = 7

We know a quadratic equation can be written as :

x² - (α + β)x + αβ

⇒ x² - (-3)x + 7

⇒ x² + 3x + 7

⇒ x² + 3x + 7 = 0 $\checkmark$

Therefore,

Quadratic equation : x² + 3x + 7 = 0 (Option : B)

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Some more informations :

* General form of quadratic equation : ax² + bx + c

* Sum of zeros = -b/a = -coefficient of x/coefficient of x²

* Product of zeros = c/a = constant term/coefficient of x²

Answer 2

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Quadratic Equations has been used. According to this, the value of variable in the equation has two values. These values are the coordinates of the points of intersection of x - axis. Using thos concept, let's do it !!

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★ Question :-

Determine the quadratic equation given the sun of the roots -3 and product of the roots 7.

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★ Solution :-

Given,

» Sum of the roots = -3

» Product of the roots = 7

• Let the required quadratic equation be

ax² + bx + c = 0

Here coefficient of x² = a, x = b and c is the constant term.

So its roots will be α and β.

We know that,

$\: \: \large{\boxed{\rm{\leadsto \: \: \alpha \: + \beta \: = \: \dfrac{(-b)}{a}}}}$

$\: \: \large{\boxed{\rm{\leadsto \: \: \alpha \: \times \: \beta \: = \: \dfrac{c}{a}}}}$

Then, according to the question, :-

~ Case I :-

⌬ α + β = -3

$\: \: \large{\rm{\longmapsto \: \: \dfrac{(-b)}{a} \: = \: \dfrac{(-3)}{1}}}$

On comparing, LHS and RHS, we get,

⌬ (-b) = (-3) and (a) = 1

Then,

$\: \: \large{\boxed{\boxed{\bf{a \: = \: 1 \: , \: b \: = \: 3}}}}$

~ Case II :-

⌬ αβ = 7

$\: \: \large{\rm{\longmapsto \: \: \dfrac{c}{a} \: = \: \dfrac{7}{1}}}$

On comparing, LHS and RHS, we get,

⌬ c = 7 and a = 1

$\: \: \large{\boxed{\boxed{\bf{c \: = \: 7}}}}$

So,

⌬ a = 1

⌬ b = 3

⌬ c = 7

Now by applying these values, in the quadratic equation we get,

⌬ x² + 3x + 7 = 0

So the correct option is Option B.) x² + 3x + 7 = 0

$\: \: \overbrace{\underbrace{\boxed{\sf{Hence, \: the \: required \: quadratic \: equation \: is \: \bf{x^{2} \: + \: 3x \: + \: 7 \: = \: 0}}}}}$

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$\large{\underline{\underline{\implies{Confused? \: Don't \: worry \: let's \: verify \: it \: :-}}}}$

For verification, we can simply apply the values we got into the Equation and check is linearity.

We know, that when we apply the value of roots, the sum should come to be 0. Then,

=> ax² + bx + c = 0

=> x² + 3x + 7 = 0

Let the value of x be k, then

=> k² + 3k + 7 = 0

Here the linearity satisfies, since we get the same value after applying the K (constant). So our answer is correct.

Hence, Verified.

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$\large{\underbrace{\tt{\leadsto \: \: Reference \: as \: Supplementary \: Counsel \: :-}}}$

• Quadratic Equations are the equations which has highest degree of variable to 2 and gives two solutions of the equations. It intersects the graph at x - axis twice.

• Different types of Polynomials are :-

• Linear Polynomial - Gives single solution and intersects x - axis just once. Example : Linear Equations

• Quadratic Polynomial - Gives two solutions.

• Cubic Polynomial - Gives three solutions and intersects x - axis thrice.

• Bi - Quadratic Polynomial - Gives four solutions and intersects x - axis four times.