QUESTION

# Find the quadratic equation whose zeroes are 7 and -5.

Hint:- Let us assume a general quadratic equation that is $a{x^2} + bx + c$ and find the value of a, b and c using the formula for sum and product of zeros of quadratic equation.

Let the quadratic equation zeros are 7 and – 5 be,
$a{x^2} + bx + c = 0$ .. (1)
Now we know that if for any quadratic equation $a{x^2} + bx + c$. If $\alpha$and $\beta$ are the zeros of this quadratic equation then,
Sum of $\alpha$and $\beta$ = $\alpha$ + $\beta$ = $- \dfrac{{{\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$ = $- \dfrac{b}{a}$
And product of $\alpha$and $\beta$ = $\alpha \times \beta$ = $\dfrac{{{\text{Constant term}}}}{{{\text{Coefficient of }}{x^2}}}$ = $\dfrac{c}{a}$
And here we are given with the value of $\alpha$and $\beta$ that is 7 and – 5.
So, now sum of zeros of the quadratic equation at equation 1 will be equal to 7 + ( – 5) = 2
So, $- \dfrac{b}{a}$ = 2 or $\dfrac{b}{a}$ = - 2 ...(2)
Now product of zeros of the quadratic equation at equation 1 will be equal to 7*( – 5) = –35
So, $\dfrac{c}{a}$ = –35 ...(3)
Now divide both sides of equation 1 by a. We get,
$\dfrac{a}{a}{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0$
${x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0$ (4)
Now to find the quadratic equation we had to put the value of $\dfrac{b}{a}$ and $\dfrac{c}{a}$ from equation 2 and 3 to equation 4. We get,
${x^2} - \left( 2 \right)x + \left( { - 35} \right) = 0$
${x^2} - 2x - 35 = 0$
Hence, the quadratic equation whose zeros are 7 and – 5 will be ${x^2} - 2x - 35 = 0$.
Note:- Whenever we come up with this type of problem then first, we had to assume that the required equation is $a{x^2} + bx + c$ and then we had to find the value of a, b and c using sum of zeros and product of zeros and after that we divide the quadratic equation by a and put the value of $\dfrac{b}{a}$ and $\dfrac{c}{a}$ in the quadratic equation to get the required quadratic equation.