Question

If 8, 2 are the roots of x2+ax+β=0 and 3, 3 are the roots of x2+αx+b=0 then the roots of the equation x2+ax+b=0 are

Answer 1

Topic

Vieta's formulas.

Solution

In a quadratic equation $ax^{2}+bx+c=0$,

the sum of roots: $-\dfrac{b}{a}$. $\bold{[1]}$

the product of roots: $\dfrac{c}{a}$. $\bold{[2]}$

From the first equation,

the sum of roots: $-\dfrac{a}{1} =8+2=10$.

From the second equation,

the product of roots: $\dfrac{b}{1} =3\cdot 3=9$.

Hence $a=-10$ and $b=9$.

From the equation,

$x^{2}-10x+9=0$

$\implies (x-1)(x-9)=0$

The roots of the equation $x^{2}+ax+b=0$ is $\boxed{{x=1} \ \textrm{or} \ x=9}$.

More information

About Vieta's formulas.

$\bold{[1]}$ $\bold{[2]}$ Vieta's formulas states that we can find the relation of the roots, like sum or product by comparing the coefficients. For example, let's take the quadratic equation that has $x=\alpha$ or $x=\beta$ as solutions, which is $ax^{2}+bx+c=0$. By factor theorem $(x-\alpha )(x-\beta )=0$.

This equation is essentially the same as the given equation. The quadratic equation should have a leading coefficient of 1, so $x^{2}+\dfrac{b}{a} x+\dfrac{c}{a} =0$.

Comparing these,

• $x^{2}+\dfrac{b}{a} x+\dfrac{c}{a} =0$
• $x^{2}-(\alpha +\beta )x+\alpha \beta =0$

So, $\boxed{\alpha +\beta =-\dfrac{b}{a}}$ and $\boxed{\alpha \beta =\dfrac{c}{a}}$.

Also, this mechanism is true for polynomials with higher degrees.

Answer 2

Given :-

If 8, 2 are the roots of x² + ax +β=0 and 3, 3 are the roots of x² + αx + b=0

To Find :-

Roots

Solution :-

For the first equation

Roots = 8 and 2

Sum = -(8 + 2)

Sum = -(10)

Sum = -10

For the second equation

Roots = 3 and 3

Product = 3 × 3

Product = 9

Equation formed = x² - 10x + 9

x² - (9x + x) + 9 = 0

x² - 9x - x + 9 = 0

x(x - 9) - 1(x - 9) = 0

(x - 9)(x - 1) = 0

Either

x - 9 = 0

x = 0 + 9

x = 9

or

x - 1 = 0

x = 0 + 1

x = 1

Roots = 9,1