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Question: Find the zeroes of the following quadratic polynomials and verify the relationship Between the ze...

Find the zeroes of the following quadratic polynomials and verify the relationship
Between the zeroes and the coefficients.
(i)x2−2x−8 (ii)4s2−4s+1(iii)6x2−3−7x (iv)4u2+8u(v)t2−15(vi)3x2−x−4  \left( {\text{i}} \right){x^2} - 2x - 8{\text{ }}\left( {{\text{ii}}} \right)4{s^2} - 4s + 1\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x \\\ \left( {{\text{iv}}} \right)4{u^2} + 8u\left( {\text{v}} \right){t^2} - 15\left( {{\text{vi}}} \right)3{x^2} - x - 4 \\\

Explanation

Solution

Hint:-To find zeros of quadratic polynomial first you have to make it in factor form and then you can find it’s zeroes and to verify the relationship between the zeroes and the coefficients use sum of zeroes is −ba\frac{{ - {\text{b}}}}{a}and product of zeroes ca\frac{{\text{c}}}{a}.

(i)x2−2x−8\left( {\text{i}} \right){x^2} - 2x - 8
To convert in factor form we will write it as
x2−4x+2x−8=0 x(x−4)+2(x−4)=0 (x−4)(x+2)=0  {x^2} - 4x + 2x - 8 = 0 \\\ x\left( {x - 4} \right) + 2\left( {x - 4} \right) = 0 \\\ \left( {x - 4} \right)\left( {x + 2} \right) = 0 \\\
Now this equation is in it’s factor form so,
x=4,x=−2x = 4,x = - 2
Now we have to verify the relationship between the zeroes and the coefficients.
Sum of zeroes is equal to −ba\frac{{ - b}}{a} on comparing with ax2+bx+c=0a{x^2} + bx + c = 0, we get (b=−2,a=1)\left( {b = - 2,a = 1} \right)
That means −ba=2\frac{{ - b}}{a} = 2and sum of zeroes (4+(−2)=2)\left( {4 + \left( { - 2} \right) = 2} \right)
∵\because Both are equal, hence the relationship is verified.
Now, product of zeroes is equal to ca\frac{c}{a} (∵c=−8,a=1)\left( {\because c = - 8,a = 1} \right)
Product of zeroes is (4×−2=−8)\left( {4 \times - 2 = - 8} \right)same as −ca=−8\frac{{ - c}}{a} = - 8, both are equal hence verified.
(ii)4s2−4s+1=0\left( {{\text{ii}}} \right)4{s^2} - 4s + 1 = 0
To convert it in factor form we will write it as
4s2−2s−2s+1=0 2s(2s−1)−1(2s−1)=0 (2s−1)(2s−1)=0  4{s^2} - 2s - 2s + 1 = 0 \\\ 2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right) = 0 \\\ \left( {2s - 1} \right)\left( {2s - 1} \right) = 0 \\\
Now this is a factor form so we can easily find ssfrom here
s=12s = \frac{1}{2} here both zeroes are same that is 12\frac{1}{2}
Now we have to verify the relationship between zeroes and the coefficients.
Here −ba=44=1\frac{{ - b}}{a} = \frac{4}{4} = 1and sum of zeroes is 12+12=1\frac{1}{2} + \frac{1}{2} = 1 both are equal hence verified.
Here ca=14\frac{c}{a} = \frac{1}{4}and product of zeroes is 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}both are equal hence verified.
(iii)6x2−3−7x=0\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x = 0
Now we have to convert it in factor form, so we will write it as
6x2−9x+2x−3=0 3x(2x−3)+1(2x−3)=0 (2x−3)(3x+1)=0  6{x^2} - 9x + 2x - 3 = 0 \\\ 3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right) = 0 \\\ \left( {2x - 3} \right)\left( {3x + 1} \right) = 0 \\\
So x=32,x=−13x = \frac{3}{2},x = \frac{{ - 1}}{3}
Now we have to verify the relationship between zeroes and coefficients.
Here −ba=76\frac{{ - b}}{a} = \frac{7}{6}and sum of zeroes 32+−13=76\frac{3}{2} + \frac{{ - 1}}{3} = \frac{7}{6}both are equal hence verified.
Here ca=−12\frac{c}{a} = \frac{{ - 1}}{2}and product of zeroes 32×−13=−12\frac{3}{2} \times \frac{{ - 1}}{3} = \frac{{ - 1}}{2}both are equal hence verified.
(iv)4u2+8u=0\left( {{\text{iv}}} \right)4{u^2} + 8u = 0
Now we have to convert it in factor form
4u(u+2)=0 ∴u=−2,u=0  4u\left( {u + 2} \right) = 0 \\\ \therefore u = - 2,u = 0 \\\
Here −ba=−2\frac{{ - b}}{a} = - 2and sum of zeroes is −2+0=−2 - 2 + 0 = - 2both are the same hence verified.
Here ca=0\frac{c}{a} = 0and product of zeroes is −2×0=0 - 2 \times 0 = 0both are the same hence verified.
(v)t2−15=0\left( v \right){t^2} - 15 = 0
We have to convert it in factor form
t2−(15)2=0 (t−15)(t+15)=0  {t^2} - {\left( {\sqrt {15} } \right)^2} = 0 \\\ \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right) = 0 \\\
t=15,t=−15t = \sqrt {15} ,t = \sqrt { - 15}
Here −ba=0\frac{{ - b}}{a} = 0 and sum of zeroes 15−15=0\sqrt {15} - \sqrt {15} = 0both are the same hence verified.
Here ca=−15\frac{c}{a} = - 15and product of zeroes 15×−15=−15\sqrt {15} \times \sqrt { - 15} = - 15both are the same hence verified.
(vi)3x2−x−4\left( {{\text{vi}}} \right)3{x^2} - x - 4
We will convert it in factor form
3x2−x−4=0 3x2−4x+3x−4=0 3x(x+1)−4(x+1)=0 (x+1)(3x−4)=0 x=−1,x=43  3{x^2} - x - 4 = 0 \\\ 3{x^2} - 4x + 3x - 4 = 0 \\\ 3x\left( {x + 1} \right) - 4\left( {x + 1} \right) = 0 \\\ \left( {x + 1} \right)\left( {3x - 4} \right) = 0 \\\ x = - 1,x = \frac{4}{3} \\\
Here −ba=13\frac{{ - b}}{a} = \frac{1}{3}and sum of zeroes 43−1=13\frac{4}{3} - 1 = \frac{1}{3}both are same hence verified.
Here ca=−43\frac{c}{a} = \frac{{ - 4}}{3}and product of zeroes 43×−1=−43\frac{4}{3} \times - 1 = \frac{{ - 4}}{3} both are same hence verified.

Note:-Whenever you get this type of question the key concept of solving is first you have to make a factor form using you basic mathematics and then using properties of quadratic equation you have to check sum of roots or zeros or product of roots with coefficients of polynomial.