
Find the quadratic equation whose zeroes are 7 and -5.
Answer
616.5k+ views
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.
Complete step-by-step answer:
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.
Complete step-by-step answer:
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.
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