If a ß are the roots of the quadratic equation x² bx c=0 then the equation whose roots are b and c

If α, β are roots of the equation ax 2+ bx + c =0 then the quadratic equation whose roots are 1/ a α+ b 2, 1/ a β+ b 2, isA. a2 x2+2 a c b2 x+1=0B. c2 x2+2 a c b2 x+1=0C. a2 c2 x2+2 a c b2 x+1=0D. a2 c2 x2+2 a c+b2 x+1=0

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We will learn the formation of the quadratic equation whose roots are given.

To form a quadratic equation, let α and β be the two roots.

Let us assume that the required equation be ax\(^{2}\) + bx + c = 0 (a ≠ 0).

According to the problem, roots of this equation are α and β.

Therefore,

α + β = - \(\frac{b}{a}\) and αβ = \(\frac{c}{a}\).

Now, ax\(^{2}\) + bx + c = 0

⇒ x\(^{2}\) + \(\frac{b}{a}\)x + \(\frac{c}{a}\) = 0 (Since, a ≠ 0)

⇒ x\(^{2}\) - (α + β)x + αβ = 0, [Since, α + β = -\(\frac{b}{a}\) and αβ = \(\frac{c}{a}\)]

⇒ x\(^{2}\) - (sum of the roots)x + product of the roots = 0

⇒ x\(^{2}\) - Sx + P = 0, where S = sum of the roots and P = product of the roots ............... (i)

Formula (i) is used for the formation of a quadratic equation when its roots are given.

For example suppose we are to form the quadratic equation whose roots are 5 and (-2). By formula (i) we get the required equation as

x\(^{2}\) - [5 + (-2)]x + 5 ∙ (-2) = 0

⇒ x\(^{2}\) - [3]x + (-10) = 0

⇒ x\(^{2}\) - 3x - 10 = 0

Solved examples to form the quadratic equation whose roots are given:

1. Form an equation whose roots are 2, and - \(\frac{1}{2}\).

Solution:

The given roots are 2 and -\(\frac{1}{2}\).

Therefore, sum of the roots, S = 2 + (-\(\frac{1}{2}\)) = \(\frac{3}{2}\)

And tghe product of the given roots, P = 2 ∙ -\(\frac{1}{2}\) = - 1.

Therefore, the required equation is x\(^{2}\) – Sx + p

i.e., x\(^{2}\) - (sum of the roots)x + product of the roots = 0

i.e., x\(^{2}\) - \(\frac{3}{2}\)x – 1 = 0

i.e, 2x\(^{2}\) - 3x - 2 = 0

2. Find the quadratic equation with rational coefficients which has \(\frac{1}{3 + 2√2}\) as a root.

Solution:

According to the problem, coefficients of the required quadratic equation are rational and its one root is \(\frac{1}{3 + 2√2}\) = \(\frac{1}{3 + 2√2}\) ∙ \(\frac{3 - 2√2}{3 - 2√2}\) = \(\frac{3 - 2√2}{9 - 8}\) = 3 - 2√2.

We know in a quadratic with rational coefficients irrational roots occur in conjugate pairs).

Since equation has rational coefficients, the other root is 3 + 2√2.

Now, the sum of the roots of the given equation S = (3 - 2√2) + (3 + 2√2) = 6

Product of the roots, P = (3 - 2√2)(3 + 2√2) = 3\(^{2}\) - (2√2)\(^{2}\) = 9 - 8 = 1

Hence, the required equation is x\(^{2}\) - Sx + P = 0 i.e., x\(^{2}\) - 6x + 1 = 0.

2. Find the quadratic equation with real coefficients which has -2 + i as a root (i = √-1).

Solution:

According to the problem, coefficients of the required quadratic equation are real and its one root is -2 + i.

We know in a quadratic with real coefficients imaginary roots occur in conjugate pairs).

Since equation has rational coefficients, the other root is -2 - i

Now, the sum of the roots of the given equation S = (-2 + i) + (-2 - i) = -4

Product of the roots, P = (-2 + i)(-2 - i) = (-2)\(^{2}\) - i\(^{2}\) = 4 - (-1) = 4 + 1 = 5

Hence, the required equation is x\(^{2}\) - Sx + P = 0 i.e., x\(^{2}\) - 4x + 5 = 0.

11 and 12 Grade Math 

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We have seen that, in the case when a parabola crosses the \(x\)-axis, the \(x\)-coordinate of the vertex lies at the average of the intercepts. Thus, if a quadratic has two real roots \(\alpha, \beta\), then the \(x\)-coordinate of the vertex is \(\dfrac{1}{2}(\alpha+\beta)\). Now we also know that this quantity is equal to \(-\dfrac{b}{2a}\). Thus we can express the sum of the roots in terms of the coefficients \(a,b,c\) of the quadratic as \(\alpha+\beta = -\dfrac{b}{a}\).

In the case when the quadratic does not cross the \(x\)-axis, the corresponding quadratic equation \(ax^2+bx+c=0\) has no real roots, but it will have complex roots (involving the square root of negative numbers). The formula above, and other similar formulas shown below, still work in this case.

We can find simple formulas for the sum and product of the roots simply by expanding out. Thus, if \(\alpha, \beta\) are the roots of \(ax^2+bx+c=0\), then dividing by \(a\) we have

Comparing the first and last expressions we conclude that

From these formulas, we can also find the value of the sum of the squares of the roots of a quadratic without actually solving the quadratic.

Suppose \(\alpha, \beta\) are the roots of \(2x^2-4x+7=0\). Find the value of \(\alpha^2+ \beta^2\) and explain why the roots of the quadratic cannot be real.

Solution

Using the formulas above, we have \(\alpha+\beta = 2\) and \(\alpha\beta = \dfrac{7}{2}\). Now \((\alpha+\beta)^2 = \alpha^2+\beta^2 + 2\alpha\beta\), so \(\alpha^2+ \beta^2 = 2^2-2\times \dfrac{7}{2}= -3\). If the roots were real, then the sum of their squares would be positive. Since the sum of their squares is \(-3\), the roots cannot be real.

Find the monic quadratic with roots \(2-3\sqrt{5}\), \(2+3\sqrt{5}\).

Suppose \(\alpha, \beta\) are the roots of \(2x^2-4x+7=0\).

Find the value of

• a \(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\)
• b \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}\).

Note that, in the previous exercise, the desired expressions are symmetric. That is, interchanging \(\alpha\) and \(\beta\) does not change the value of the expression. Such expressions are called symmetric functions of the roots.

Suppose \(\alpha, \beta\) are the roots of \(ax^2+bx+c=0\). Find a formula for \((\alpha - \beta)^2\) in terms of \(a\), \(b\) and \(c\).

You may recall that the arithmetic mean of two positive numbers \(\alpha\) and \(\beta\) is \(\dfrac{1}{2}(\alpha + \beta)\), while their geometric mean is \(\sqrt{\alpha\beta}\). Thus, if a quadratic has two positive real solutions, we can express their arithmetic and geometric mean using the above formulas.

Without solving the equation, find the arithmetic and geometric mean of the roots of the equation \(3x^2-17x+12=0\).

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