
A quadratic equation whose one root is 3 is
(a) ${{z}^{2}}-6z-5=0$
(b) ${{z}^{2}}+6z-5=0$
(c) ${{z}^{2}}-5z-6=0$
(d) ${{z}^{2}}-5z+6=0$
Answer
487.8k+ views
Hint: The only way to solve this question is through option verification. It is given that the root is 3. This means, if we substitute z = 3 in the quadratic equations, the left-hand side must be equal to the right-hand side. We will consider each option one by one and substitute z = 3. If the left-hand side is equal to the right-hand side, that option is the correct option.
Complete step-by-step solution:
Let us first consider the option (a). We will substitute z = 3 to see whether the lefthand side is equal to the right-hand side or not.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-6\left( 3 \right)-5$
$\Rightarrow $ LHS = 9 – 18 – 5
$\Rightarrow $ LHS = ─24
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (a) does not verify.
Now, let us consider option (b) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}+6\left( 3 \right)-5$
$\Rightarrow $ LHS = 9 + 18 – 5
$\Rightarrow $ LHS = 22
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (b) does not verify.
Now, we will consider option (c) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)-6$
$\Rightarrow $ LHS = 9 – 15 – 6
$\Rightarrow $ LHS = ─12
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (c) does not verify.
The option left with us is an option (d) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)+6$
$\Rightarrow $ LHS = 9 – 15 + 6
$\Rightarrow $ LHS = 0
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS = RHS
Thus, option (d) verify with the conditions.
Hence, option (d) is the correct option.
Note: The other method is to actually solve the equations and find the roots of that equation. Those equations can be solved with the factorization method, completing the square method or formula method. But if it is possible to verify the equations, we shall opt for verification as it is quicker and relatively simple.
Complete step-by-step solution:
Let us first consider the option (a). We will substitute z = 3 to see whether the lefthand side is equal to the right-hand side or not.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-6\left( 3 \right)-5$
$\Rightarrow $ LHS = 9 – 18 – 5
$\Rightarrow $ LHS = ─24
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (a) does not verify.
Now, let us consider option (b) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}+6\left( 3 \right)-5$
$\Rightarrow $ LHS = 9 + 18 – 5
$\Rightarrow $ LHS = 22
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (b) does not verify.
Now, we will consider option (c) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)-6$
$\Rightarrow $ LHS = 9 – 15 – 6
$\Rightarrow $ LHS = ─12
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS $\ne $ RHS
Thus, option (c) does not verify.
The option left with us is an option (d) and substitute z = 3.
$\Rightarrow $ LHS = ${{\left( 3 \right)}^{2}}-5\left( 3 \right)+6$
$\Rightarrow $ LHS = 9 – 15 + 6
$\Rightarrow $ LHS = 0
$\Rightarrow $ RHS = 0
$\Rightarrow $ LHS = RHS
Thus, option (d) verify with the conditions.
Hence, option (d) is the correct option.
Note: The other method is to actually solve the equations and find the roots of that equation. Those equations can be solved with the factorization method, completing the square method or formula method. But if it is possible to verify the equations, we shall opt for verification as it is quicker and relatively simple.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

State the laws of reflection of light
