Question

If one root of quadratic equation ax²+bx+c=0 is 3-4i, then a+b+c=​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ ax² + bx + c.

One roots of quadratic equation = 3 - 4i.

As we know that,

Let one roots be = 3 - 4i.

Other roots be = 3 + 4i.

Sum of the zeroes of the quadratic equation.

⇒ α + β = - b/a.

⇒ 3 + 4i + 3 - 4i = 6.

Products of zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ (3 - 4i) x (3 + 4i).

As we know that,

Formula of :

⇒ (x² - y²) = (x + y)(x - y).

Put the values in the equation, we get.

⇒ [(3)² - (4i)²]

⇒ 9 - 16i².

⇒ 9 - 16(-1) = 25.

⇒ αβ = 25.

As we know that,

Formula of quadratic equation.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (6)x + 25.

⇒ x² - 6x + 25.

As we can see that,

⇒ a = 1  and  b = -6  and  c = 25.

Value of : a + b + c.

⇒ 1 - 6 + 25 = 20.

a + b + c = 20.

                                                                                                                     

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Answer:

Given :-

• If one of quadratic equation
• ${ax}^{2} + bx + c \: \: \: is \: 3 - 4i$

To find :-

• The value of
• a+b+c=?

Explanation :-

• As we know from the question that,
• one root of the quadratic equation =3-4i.

Let,

• One root of quadratic equation =3-4i
• Other root of quadratic equation =3+4i

• We know that,

• Sum of the zeroes of quadratic equation is

• $\alpha + \beta = \frac{ - b}{a}$
• Applying the values,

• $3 + 4i + 3 - 4i = 6$

And,

• Product of zeroes Of quadratic equation

• $\alpha \beta = \frac{c}{a}$
• Apply the values,

• $(3 - 4i) \times (3 + 4i)$

Here,

• We know one formula that

• $( {x}^{2} - {y}^{2} ) = (x + y)(x - y)$
• Now lets put the value in the equation we get that,

• $(( {3)}^{2} - (4 {i)}^{2} )$

• $9 - 16 {i}^{2}$

• $9 - 16( - 1) = 25. \: \: \: \: \alpha \beta = 25$

• We know the formula of quadratic equation that is ,

• ${x}^{2} - ( \alpha + \beta )x + \alpha \beta$
• Now applying all the values we get that,

• ${x}^{2} - 6(x) + 25$

• ${x}^{2} - 6x + 25$
• Here we can easily see that,
• value of a=1
• Value of b=-6
• Value of c=25.

♧According to the given question,

• a+b+c=1+(-6)+25=20.

Hope it helps u mate .

Thank you .